RD Sharma Solutions Class 10 Arithmetic Progressions Exercise 9.6

RD Sharma Class 10 Solutions Chapter 9 Ex 9.6 PDF Free Download

Exercise 9.6

 

1. Find the sum of the following Arithmetic Progression.

(i) 50, 46, 42,… To 10 terms

(ii) 1, 3, 4, 7, . . . 26 to 12 terms.

(iii) \(3, \frac{9}{2}, 6, \frac{15}{2}, . . . .\) to 25 terms.

(iv) 41, 36, 31, . . .  To 12 terms.

(v) a + b, a – b, a -3b,… To 22 terms.

(vi) (x – y)2 , (x2, y2), (x + y)2, . . . to n terms.

(vii) \(\frac{\left ( x – y \right )}{\left ( x + y \right )}, \frac{\left (3x – 2y \right )}{\left ( x + y \right )}, \frac{\left ( 5x – 3y \right )}{\left ( x + y \right )}, . . . to n terms\)

(viii) -26, -24, -22, . . . to 36 terms.

Solution:-

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

Then, According to the question,

(i) 50, 46, 42,… To 10 terms

Common difference of the A.P. (d)

= a2 – a1

= 46 – 50

= -4

Number of terms (n) = 10

First term for the given A.P. (a) = 50

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Then, according to the question,

\(S_{10} = \frac{10}{2}\left [ 2(50) + \left ( 10 – 1 \right )\left ( -4 \right )\right ]\)

= (5) [ 100 + (9)(-4) ]

= (5) [ 100 – 36 ]

= (6) [64]

= 320

∴, the sum of first 10 terms of the given A.P. is 320

 

(ii) 1, 3, 4, 7, . . . 26 to 12 terms.

Common difference of the A.P. (d)

= a2 – a1

= 3 – 1

= 2

Number of terms (n) = 12

First term (a) = 1

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Then, according to the question,

\(S_{12} = \frac{12}{2}\left [ 2(1) + \left ( 12 – 1 \right )\left ( 2 \right )\right ]\)

= (6) [ 2 + (11)(2) ]

= (6) [ 2 + 22 ]

= (6) [24]

= 144

∴, the sum of first 10 terms of the given A.P. is 144

 

(iii) \(3, \frac{9}{2}, 6, \frac{15}{2}, . . . .\) to 25 terms.

Common difference here is (d): a2 – a1

= \(\frac{9}{2} – 3\)

= \(\frac{9 – 6}{2}\)

= \(\frac{3}{2}\)

Number of terms (n) = 25

First term of the A.P. (a) = 3

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Then, according to the question,

\(S_{25} = \frac{25}{2}\left [ 2(3) + \left ( 25 – 1 \right )\left (\frac{3}{2}  \right )\right ]\)

= \(\left ( \frac{25}{2} \right )\left [ 6 + \left ( 24\right )\left ( \frac{3}{2} \right ) \right ]\)

= \(\left ( \frac{25}{2} \right )\left [ 6 + \left ( \frac{72}{2}\right )\right ]\)

= \(\left ( \frac{25}{2} \right )\left [ 6 + 36\right ]\)

= \(\left ( \frac{25}{2} \right )\left [ 42\right ]\)

= ( 25 ) ( 21 )

= 525

∴, the sum of first 12 terms for the given A.P. is 162.

 

(iv) 41, 36, 31, . . .  To 12 terms.

Common Difference of the A.P. (d) = a2 – a1

= 36 – 41

= -5

Number of terms (n) = 12

First term for the given A.P. (a) = 41

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Then, according to the question,

\(S_{12} = \frac{12}{2}\left [ 2\left ( 41 \right ) + \left ( 12 – 1\right )\left ( -5 \right ) \right ]\)

= (6) [82 + (11) (-5)]

= (6) [82 – 55]

= (6) [27]

=162

∴, the sum of first 12 terms for the given A.P. 162

 

(v) a + b, a – b, a -3b,… To 22 terms.

Common difference of the A.P. (d) = a2 – a1

= (a –  b) – (a + b)

=a – b – a – b

= -2b

Number of terms (n) = 22

First term for the given A.P. (a) = a + b

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Then, according to the question,

\(S_{22} = \frac{22}{2}\left [ 2\left ( a + b \right ) + \left ( 22 – 1 \right )\left ( -2b \right ) \right ]\)

= (11)[ 2(a + b) + (22 – 1)( -2b ) ]

= (11)[ 2a + 2b + (21)(-2b) ]

= (11)[ 2a + 2b – 42b ]

= (11)[2a – 40b]

= 22a – 40b

∴, the sum of first 22 terms for the given A.P. is: 22a – 40b

 

(vi) (x – y)2 , (x2, y2), (x + y)2, . . . to n terms.

Common difference of the A.P. (d) = a2 – a1

= (x2 – y2) – (x – y)2

= x2 + y2 – (x2 + y2 – 2xy)

= 2xy

Number of terms (n) = n

First term for the given A.P. (a) = (x – y)2

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Then, according to the question,

\(S_{n} = \frac{n}{2}\left [ 2\left ( x – y \right )^{2} + \left ( n – 1 \right )2xy \right ]\)

= \(\frac{n}{2}\left ( 2 \right )\left[\left ( x – y \right )^{2} + \left ( n – 1 \right ) xy \right ]\)

= (n)[ ( x – y )2 + ( n – 1 )xy ]

∴, the sum of first n terms of the given A.P. is (n)[ ( x – y )2 + ( n – 1 )xy ]

  

(vii) \(\frac{\left ( x – y \right )}{\left ( x + y \right )}, \frac{\left (3x – 2y \right )}{\left ( x + y \right )}, \frac{\left ( 5x – 3y \right )}{\left ( x + y \right )}, . . . to n terms\)

Common difference of the A.P. (d) = a2 – a1

= \(\left ( \frac{3x – 2y}{x + y} \right ) – \left ( \frac{x – y}{ x + y} \right )\)

= \(\frac{\left ( 3x – 2y \right ) – \left ( x – y \right )}{x + y}\)

= \(\frac{3x – 2y – x + y}{x + y}\)

= \(\frac{2x – y}{x + y}\)

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Then, according to the question,

\(S_{n} = \frac{n}{2}\left [ 2\left ( \frac{2x – 2y}{x + y}\right ) + \left ( n – 1\right )\left ( \frac{2x – y}{x + y} \right ) \right ]\)

= \(\left ( \frac{n}{2} \right )\left [ \left ( \frac{2x – 2y}{x + y} \right ) + \left ( \frac{\left ( n – 1 \right )\left ( 2x – y \right )}{x + y} \right ) \right ]\)

= \(\left ( \frac{n}{2} \right )\left [ \left ( \frac{2x – 2y}{x + y} \right ) + \left ( \frac{n\left ( 2x – y \right ) – 1\left ( 2x – y \right )}{x + y} \right ) \right ]\)

= \(\left ( \frac{n}{2} \right )\left ( \frac{2x – 2y + n\left ( 2x – y \right ) -2x + y}{x + y} \right )\)

= \(\left ( \frac{n}{2} \right )\left ( \frac{n\left ( 2x – y \right ) – y}{x + y} \right )\)

∴, the sum of first n terms for the given A.P. is \(\left ( \frac{n}{2} \right )\left ( \frac{n\left ( 2x – y \right ) – y}{x + y} \right )\)

 

(viii) -26, -24, -22, . . . to 36 terms.

Common difference of the A.P. (d) = a2 – a1

= (-24) – (-26)

= -24 + 26

=2

Number of terms (n) = 36

First term for the given A.P. (a) = -26

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Then, according to the question,

\(S_{n} = \left ( \frac{36}{2} \right )\left [ 2 \left ( -26 \right ) + \left ( 36 – 1 \right )\left ( 2 \right ) \right ]\)

= (18) [ -52 + (35) (2)

= (18) [-52 + 70]

= (18) (18)

= 324

∴, the sum of first 36 terms for the given A.P. is 324

 

2. Find the sum to n term of the A.P. 5, 2, -1, -4, -7, . . .

Solution:

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Then, according to the question,

\(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term for the given A.P.

d = common difference of the given A.P.

and n = number of terms

For the given A.P. (5,2,-1,-4,-7,…),

Common difference of the A.P. (d) = a2 – a1

= 2 – 5

= -3

Number of terms (n) = n

First term for the given A.P. (a) = 5

Substituting on the formula, we get,

\(S_{n} = \frac{n}{2}\left [ 2\left ( 5 \right ) + \left ( n – 1 \right ) \left ( – 3 \right ) \right ]\)

= \(\frac{n}{2}\left [ 10 + \left ( -3n + 3 \right ) \right ]\)

= \(\frac{n}{2}\left [ 10 – 3n + 3 \right ]\)

= \(\frac{n}{2}\left [ 13 – 3n \right ]\)

∴, the sum of first n terms for the given A.P. is \(\frac{n}{2}\left [ 13 – 3n \right ]\)

  

 

3. Find the sum of n terms of an A.P. whose nth terms is given by an = 5 – 6n.

Solution: 

nth term of the A.P.=an = 5 – 6n

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Then, according to the question,

Where, a = the first term

l = the last term

So, for the given AP,

The first term (a) will be calculated using n = 1 in the given equation for nth term of A.P.

a = 5 – 6(1)

= 5 -6

= -1

Now, the last term (l) or the nth term is given

an = 5 – 6n

Substituting in the formula, we get,

\(S_{n} = \left ( \frac{n}{2} \right )\left [ \left ( -1 \right ) + 5 -6n \right ]\)

= \(\left ( \frac{n}{2} \right )\left [ 4 – 6n \right ]\)

= \(\left ( \frac{n}{2} \right )\left ( 2 \right )\left [ 2 – 3n \right ]\)

= (n) (2 – 3n)

∴, the sum of the n terms of the given A.P. is (n) (2 – 3n)

 

5. Find the sum of first 15 term of each of the following sequences having nth term as

(i) an = 3 + 4n

(ii) bn = 5 + 2n

(iii) xn = 6 – n

(iv) yn = 9 -5n

Solution:

(i) Here, we are given an A.P. whose nth term is given by the following expression,

nth term, an = 3 + 4n

We need to find the sum of first 15 term & n ,

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where, a = the first term

l = the last term

Then, according to the question,

The first term (a) will be calculated using n = 1 in the given equation for nth term of A.P.

a = 3 + 4(1)

= 3 + 4

=7

Now, the last term (l) or the nth term is given

l = an = 3 + 4n

So, on substituting the values in the formula for the sum of n terms of an A.P, we get,

\(S_{n} = \frac{15}{2}\left ( 7 + 3 + 4(15) \right )\)

= \(\frac{15}{2}\left ( 10 + 60 \right )\)

= \( \frac{15}{2}\left ( 70 \right )\)

= (15)(35)

= 525

∴, the sum of the 15 terms of the given A.P. is S15 = 525

 

(ii) Here, we are given an A.P. whose nth term is given by the following expression,

bn = 5 + 2n

We need to find the sum of first 15 term & n ,

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where, a = the first term

l = the last term

Then, according to the question,

The first term (a) will be calculated using n = 1 in the given equation for nth term of A.P.

b = 5 + 2(1)

= 5 + 2

= 7

Now, the last term (l) or the nth term is given

l = bn = 5 + 2n

So, on substituting the values in the formula for the sum of n terms of an A.P, we get,

\(S_{n} = \frac{15}{2}\left ( 7 + 5 + 2(15) \right )\)

= \(\frac{15}{2}\left ( 12 + 30 \right )\)

= \(\frac{15}{2}\left ( 42 \right )\)

= (15) (21)

= 315

Therefore, the sum of the 15th term of the given A.P. is 315

 

(iii) Here, we are given an A.P. whose nth term is given by the following expression,

xn = 6 – n

We need to find the sum of first 15 term & n ,

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where, a = the first term

l = the last term

Then, according to the question,

The first term (a) will be calculated using n = 1 in the given equation for nth term of A.P.

b = 6 – 1

= 5

Now, the last term (l) or the nth term is given

l = xn = 6 – n

So, on substituting the values in the formula for the sum of n terms of an A.P, we get,

\(S_{n} = \frac{15}{2}\left ( (5) + 6  – (15) \right )\)

= \(\frac{15}{2}\left ( 11 – 15 \right )\)

= \(\frac{15}{2}\left ( -4 \right )\)

= (15) (-2)

= -30

Therefore, the sum of the 15 terms of the given A.P. is -30.

 

(iv) Here, we are given an A.P. whose nth term is given by the following expression,

yn = 9 – 5n

We need to find the sum of first 15 term & n ,

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where, a = the first term

l = the last term

Then, according to the question,

The first term (a) will be calculated using n = 1 in the given equation for nth term of A.P.

b = 9  – 5 (1)

= 9 – 5

= 4

Now, the last term (l) or the nth term is given

l = bn = 9 – 5n

So, on substituting the values in the formula for the sum of n terms of an A.P, we get,

\(S_{n} = \frac{15}{2}\left ( (4) + 9 – 5(15) \right )\)

= \(\frac{15}{2}\left ( 13 – 75 \right )\)

= \(\frac{15}{2}\left ( – 62 \right )\)

= (15) (-31)

= -465

Therefore, the sum of the 15 terms of the given A.P. is -465

 

 6. Find the sum of first 20 terms the sequence whose nth term is an= An + B.

Solution:

nth term, an = An + B

We need to find the sum of first 20 terms.

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where, a = the first term

l = the last term

Then, according to the question,

The first term (a) will be calculated using n = 1 in the given equation for nth term of A.P.

a = A(1) + B

= A + B

Now, the last term (l) or the nth term is given

l = an =  An + B

Substituting the values in the formula,

\(S_{20} = \frac{20}{2}\left ( \left ( A + B \right ) + A\left ( 20 \right ) + B \right )\)

= 10[ 21A + 2B ]

= 210A + 20B

∴, the sum of the first 20 terms of the given A.P. is 210A+20B

 

 

7. Find the sum of first 25 terms of an A.P whose nth term is given by an = 2 – 3n.

Solution:

nth term, anan = 2 -3n

We need to find the sum of first 25 terms.

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where, a = the first term

l = the last term

Then, according to the question,

The first term (a) will be calculated using n= 1 in the given equation for nth term of A.P.

a =2 -3(1)

=2-3

= -1

Now, the last term (1) or the nth term is given l = an =2 – 3n

Substituting the values in the formula,

\(S_{25} = \frac{25}{2}\left [ \left ( -1 \right ) + 2 – 3\left ( 25 \right ) \right ]\)

= \(\frac{25}{2}\left [ 1 – 75 \right ]\)

= (25) (-37)

= -925

∴, the sum of the 25 terms of the given A.P. is -925

 

 

8. Find the sum of first 25 terms of an A.P whose nth term is given by an = 7 – 3n.

Solution:

nth term, anan = 7 – 3n.

We need to find the sum of first 25 terms.

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where, a = the first term

l = the last term

Then, according to the question,

The first term (a) will be calculated using n= 1 in the given equation for nth term of A.P.

a = 7 – 3

= 4

Now, the last term (l) or the nth term is given

l = an = 7 – 3n

Substituting the values in the formula,

\(S_{n} = \frac{25}{2}\left [ \left ( 4 \right ) + 7 – 3\left ( 25 \right ) \right ]\)

\(\: = \frac{25}{2}\left [ 11 – 75 \right ]\)

\(\: = \frac{25}{2}\left [ -64 \right ]\)

\(\: = \left ( 25 \right ) \left ( -32 \right )\)

\(\: = -800\)

∴, the sum of the 25 terms of the given A.P. is S­n = -800

 

 

9. If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, . . . ,  is 116. Find the last term.

Solution:-  

So here, let us first find the number of terms whose sum is 116.

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So for the given A.P(25, 22, 19,…)

The first term (a) = 25

The sum of n terms Sn = 116

Common difference of the A.P. (d) = a2 – a1

= 22-25

= -3

Substituting the values in the formula,

=> \(116 = \frac{n}{2}\left [ 2\left ( 25 \right ) + \left ( n – 1 \right )\left ( – 3 \right ) \right ]\)

=> \(\left ( \frac{n}{2} \right )\left [ 50 + \left ( -3 + 3 \right ) \right ]\)

=> \(\left ( \frac{n}{2} \right )\left [ 53 – 3n \right ]\)

=> 116 x 2 = 53n – 3n2

So, we get the following quadratic equation, 3n2 – 53n + 232 = 0

On solving by splitting middle term, we get,

=> 3n2 – 24n – 29n + 232 = 0

=> 3n( n – 8 ) – 29 ( n – 8 ) = 0

=> (3n – 29) ( n – 8 ) = 0

Further,

3n – 29 = 0

=> n = \(\frac{29}{3}\)

Also,

n – 8 = 0

=> n = 8

Since, n cannot be a fraction, so the number of terms is 8.

So, the term is:

a8 = a1 + 7d

= 25 + 7(-3)

= 25 – 21

= 4

Therefore, the last term of the given A.P. such that the sum of the terms is 116 is 4.

 

  

10.

(i). How many terms of the sequence 18, 16, 14…. should be taken so that their sum is 0 (Zero).

(ii). How many terms are there in the A.P. whose first and fifth terms are  -14 and 2 respectively and the sum of the terms is 40?

(iii). How many terms of the A.P. 9, 17, 25, . . . must be taken so that their sum is 636?

(iv). How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693?

(v). How many terms of the A.P. is 27, 24, 21. . . should be taken that their sum is zero?

Solution:

(i) AP. is 18,16,14,…

So here, let us find the number of terms whose sum is 0.

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

The first term (a) = 18

The sum of n terms (Sn) = 0

Common difference of the A.P. (d) = a2  – a1

= 16 – 18

= -2

Substituting the values in the formula,

=> \(0 = \frac{n}{2}\left [ 2\left ( 18 \right ) + \left ( n – 1 \right )\left ( -2 \right ) \right ]\)

=> \(0 = \frac{n}{2}\left [ 36 + \left ( -2n + 2 \right ) \right ]\)

=> \(0 = \frac{n}{2}\left [ 38 – 2n \right ]\)

Further,

\(\frac{n}{2}\)

=> n = 0

Or, 38 – 2n = 0

=> 2n = 38

=> n = 19

Since, the number of terms cannot be zero; the number of terms (n) is 19

  

(ii) Here, let us take the common difference as d.

So, we are given,

First term (a1) = -14

Filth term (a5) = 2

Sum of terms (Sn) = 40

Now,

a5 = a1 + 4d

=> 2 = -14 + 4d

=> 2 + 14 = 4d

=> 4d = 16

=> d=4

Further, let us find the number of terms whose sum is 40.

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

The first term (a1) = -14

The sum of n terms (Sn) = 40

Common difference of the A.P. (d) = 4

Substituting the values in the formula,

=> \(40 = \frac{n}{2}\left [ 2 \left ( -14 \right ) + \left ( n – 1 \right )\left ( 4 \right ) \right ]\)

=> \(40 = \frac{n}{2}\left [ -28 + \left ( 4n – 4 \right ) \right ]\)

=> \(40 = \frac{n}{2}\left [ -32 + 4n \right ]\)

=> 40 (2) = – 32n + 4n2

So, we get the following quadratic equation,

4n2 – 32n – 80 = 0

=> n2 – 8n + 20 = 0

On solving by splitting the middle term, we get

4n2 – 10n + 2n + 20 = 0

=> n( n – 10 ) + 2( n – 10 ) = 0

=> ( n + 2 ) ( n – 10) = 0

Further,

n + 2 = 0

=> n = -2

Or,

n – 10 = 0

=> n = 10

Since the number of terms cannot be negative.

Therefore, the number of terms (n) is 10.

 

(iii) AP is 9,17,25,…

So here, let us find the number of terms whose sum is 636.

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

The first term (a) = 9

The sum of n terms (Sn) = 636

Common difference of the A.P. (d) = a2  – a1

= 17 – 9

= 8

Substituting the values in the formula,

we get,

=> \(636 = \frac{n}{2}\left [ 2\left ( 9 \right ) + \left ( n – 1 \right )\left ( 8 \right ) \right ]\)

=> \(636 = \frac{n}{2}\left [ 18 + \left ( 8n – 8 \right ) \right ]\)

=> 636 (2) = (n) [ 10 + 8n ]

=> 1271 = 10n + 8n2

So, we get the following quadratic equation,

=> 8n2 + 10n – 1272 = 0

=> 4n2 + 5n – 636 = 0

On solving by splitting the middle term, we get,

=> 4n2 – 48n + 53n – 636 = 0

=> 4n( n – 12 ) – 53( n – 12 ) = 0

=> ( 4n – 53 ) ( n – 12 ) = 0

Further,

4n – 53 = 0

=> \(n = \frac{53}{4}\)

Or, n – 12 = 0

=> n = 12

Since, the number of terms cannot be a fraction.

Therefore, the number of terms (n) is 12.

  

(iv) A.P. is 63,60,57,…

So here. let us find the number of terms whose sum is 693.

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

The first term (a) = 63

The sum of n terms (Sn) = 693

Common difference of the A.P. (d) = a2 – a1

= 60 – 63

= -3

Substituting the values in the formula,

=> \(693 = \frac{n}{2} \left [ 2\left ( 63 \right ) + \left ( n – 1 \right )\left ( -3 \right ) \right ]\)

=> \(693 = \frac{n}{2} \left [ 163 + \left ( -3n + 3 \right ) \right ]\)

=> \(693 = \frac{n}{2} \left [ 129 – 3n \right ]\)

=> 693 (2) = 129n – 3n2

So. we get the following quadratic equation.

=> 3n2 – 129n + 1386 = 0

=> n2 – 43n + 462

On solving by splitting the middle term, we get.

=> n2 – 22n – 21n + 462 = 0

=> n( n – 22 ) -21( n – 22 ) = 0

=> ( n – 22) ( n – 21 ) = 0

Further,

n – 22 = 0

=> n = 22

Or,       n – 21 = 0

=> n = 21

Here, 22nd term will be

a22 = a1 + 21d

= 63 + 21( -3 )

= 63 – 63

= 0

So, the sum of 22 as well as 21 terms is 693.

Therefore, the number of terms (n) is 21 or 22

 

(v) A.P. is 27, 24, 21. . .

So here. let us find the number of terms whose sum is 0.

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

The first term (a) = 27

The sum of n terms (Sn) = 0

Common difference of the A.P. (d) = a2 – a1

= 24 – 27

= -3

Substituting the values in the formula,

=> \(0 = \frac{n}{2}\left [ 2\left ( 27 \right ) + \left ( n – 1 \right )\left ( -3 \right ) \right ]\)

=> 0 = (n) [ 54 + ( n – 1 )(-3) ]

=> 0 = (n) [ 54 – 3n + 3 ]

=> 0 = n [ 57 – 3n ]

Further we have,

n = 0

Or,

57 – 3n = 0

=> 3n = 57

=> n = 19

The number of terms cannot be zero,

Therefore, the numbers of terms (n) is 19.

 

 

11. Find the sum of the first

(i) 11 terms of the A.P. : 2, 6, 10, 14,  . . .

(ii) 13 terms of the A.P. : -6, 0, 6, 12, . . .

(iii) 51 terms of the A.P. : whose second term is 2 and fourth term is 8.

Solution:

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

 

(i) 2,6,10,14,… To 11 terms.

Common difference of the A.P. (d) = a2 – a1

= 10 – 6

= 4

Number of terms (n) = 11

First term for the given A.P. (a) = 2

Substituting in the formula, we get,

\(S_{11} = \frac{11}{2}\left [ 2  \left ( 2 \right ) + \left ( 11 – 1 \right )4 \right ]\)

= \(\frac{11}{2}\left [ 2\left ( 2 \right ) + \left ( 10 \right )4 \right ]\)

= \(\frac{11}{2}\left [ 4 + 40 \right ]\)

= 11 X 22

= 242

Therefore, the sum of first 11 terms for the given A.P. is 242

 

(ii) -6, 0, 6, 12, … to 13 terms.

Common difference of the AR (d) = a2 – a­­1

= 6 – 0

= 6

Number of terms (n) = 13

First term for the given AP (a) = -6

Substituting in the formula, we get,

\(S_{13} = \frac{13}{2}\left [ 2  \left ( -6 \right ) + \left ( 13 – 1 \right )6 \right ]\)

= \(\frac{13}{2}\left [\left( -12 \right ) + \left ( 12 \right )6 \right ]\)

= \(\frac{13}{2}\left [ 60 \right ]\)

= 390

Therefore, the sum of first 13 terms for the given AR is 390

 

(iii) 51 terms of an AP whose a2 = 2 and a­­4 = 8

Now,

a2 = a + d

2 = a + d                                                                                              …(i)

Also,

a4 = a + 3

8 = a + 3d                                                                                            … (2)

Subtracting (1) from (2), we get

2d = 6

d = 3

Substituting d = 3 in (i), we get

2 = a + 3

=> a = -1

Number of terms (n) = 51

First terms for the given A.P.(a) = -1

Substituting in the formula, we get,

\(S_{n} = \frac{51}{2}\left [ 2 \left ( -1 \right ) + \left ( 51 – 1 \right )\left ( 3 \right ) \right ]\)

= \(\frac{51}{2}\left [ -2 + 150 \right ]\)

= \(\frac{51}{2}\left [ 158 \right ]\)

= 3774

Therefore, the sum of first 51 terms for the A.P. is 3774.

 

12. Find the sum of

(i) First 15 multiples of 8

(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

(iii) all 3 – digit natural numbers which are divisible by 13.

Solution:

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where: a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

 

(i) First 15 multiples of 8.

So, we know that the first multiple of 8 is 8 and the last multiple of 8 is 120.

Also, all these terms will form an A.P. with the common difference of 8.

So here,

First term (a) = 8

Number of terms (n) = 15

Common difference (d) = 8

Using the formula,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

\(S_{n} = \frac{15}{2}\left [ 2\left ( 8 \right ) + \left ( 15 – 1 \right )8 \right ]\)

= \( \frac{15}{2} [ 16 +  (14)8 ]\)

= \( \frac{15}{2} [16 + 12]\)

= \( \frac{15}{2} [ 128 ]\)

= 960

Therefore, the sum of the first 15 multiples of 8 is 960

 

(ii)

(a) First 40 positive integers divisible by 3

So, we know that the first multiple of 3 is 3 and the last multiple of 3 is 120.

Also, all these terms will form an A.P. with the common difference of 3.

So here,

First term (a) = 3

Number of terms (n) = 40

Common difference (d) = 3

Using the formula,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

\(S_{n} = \frac{40}{2}\left [ 2\left ( 3 \right ) + \left ( 40 – 1 \right )3 \right ]\)

= 20 [ 6 + (39)3 ]

= 20 (6 + 117)

=20 (123)

= 2460

Therefore, the sum of first 40 multiples of 3 is 2460

 

(b) First 40 positive integers divisible by 5

So, we know that the first multiple of 5 is 5 and the last multiple of 5 is 200.

Also, all these terms will form an A.P. with the common difference of 5.

So here,

First term (a) = 5

Number of terms (n) = 40

Common difference (d) = 5

Using the formula,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

\(S_{n} = \frac{40}{2}\left [ 2\left (5 \right ) + \left ( 40 – 1 \right )5 \right ]\)

= 20 [ 10 + (39)5 ]

= 20 (10 + 195)

= 20 (205)

= 4100

Therefore, the sum of first 40 multiples of 5 is 4100

 

(c) First 40 positive integers divisible by 6

So, we know that the first multiple of 6 is 6 and the last multiple of 6 is 240.

Also, all these terms will form an A.P. with the common difference of 6.

So here,

First term (a) = 6

Number of terms (n) = 40

Common difference (d) = 6

Using the formula,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

\(S_{n} = \frac{40}{2}\left [ 2\left (6 \right ) + \left ( 40 – 1 \right )6 \right ]\)

= 20[12 + (39) 6]

=20 (12 + 234)

= 20( 246 )

= 4920

Therefore, the sum of first 40 multiples of 6 is 4920

 

(ii) All 3 digit natural number which are divisible by 13

So, we know that the first 3 digit multiple of 13 is 104

and the last 3 digit multiple of 13 is 988.

Also, all these terms will form an AR with the common difference of 13.

So here,

First term (a) = 104

Last term (l) = 988

Common difference (d) = 13

So, here the first step is to find the total number of terms.

Let us take the number of terms as n.

Now, as we know,

\(a_{n} = a + \left ( n – 1 \right )d\)

So, for the last term,

988 = 104 + (n – 1)13

=> 988 = 104 + 13n – 13

=> 988 = 91 + 13n

=> 13n = 897

=> n = 69

Using the formula,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

\(S_{n} = \frac{69}{2}\left [ 2\left ( 104 \right ) + \left ( 69 – 1 \right )13 \right ]\)

= \(\frac{69}{2}\left [ 208 + 884 \right ]\)

= \(\frac{69}{2}\left [ 1092 \right ]\)

= 69 (546)

= 37674
Therefore, the sum of all 3 digit multiples of 13 is 37674

 

 

13. Find the sum:

(i) 2 + 4 + 6 + . . . + 200

(ii) 3 + 11 + 19 + . . . + 803

(iii) (-5) + (-8) + (-11) + . . . + (-230)

(iv) 1 + 3 + 5 + 7 + . . . + 199

(v) \(7 + 10\frac{1}{2} + 14 + . . . + 84\)

(vi) 34 + 32 + 30 + . . . + 10

(vii) 25 + 28 + 31 + . . . + 100

Solution:

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

or

\(S_{n} = \frac{n}{2} [ a + l ]\)

Where; a = first term of the given A.P.

d = common difference of the given A.P.

l = last term

n = number of terms

 

(i) 2 + 4 + 6 + … + 200

Common difference of the A.P. (d) = a2 – a1

= 6 – 4

= 2

So here,

First term (a) = 2

Last term (l) = 200

Common difference (d) = 2

Let us take the number of terms as n.

Now, as we know,
an = a + (n – 1)d

So, the last term,

200 = 2 + (n – 1)2

200= 2 + 2n – 2

200 = 2n

n = 100

Substituting in the formula,

\(S_{100} = \frac{100}{2}\left [ a + l \right ]\)

= 50 [ 2 + 200 ]

= 50 X 202

= 10100

Therefore, the sum of the A.P is 10100

 

(ii) 3 + 11 + 19 + . . . + 803

Common difference of the A.P. (d) = a2 – a1

=19 – 11

= 8

So here,

First term (a) = 3

Last term (l) = 803

Common difference (d) = 8

Let us take the number of terms as n.

an = a + (n – 1)d

So, the last term,

803=3 + (n – 1)8

=> 803= 3 + 8n – 8

=> 803 + 5 = 8n

=> 808 = 8n

=> n =101

Substituting in the formula,

\(S_{101} = \frac{101}{2} \left [ a + l \right ]\)

= \(\frac{101}{2} \left [ 3 + 803 \right ]\)

= \(\frac{101}{2} \left [ 806 \right ]\)

= 101 (403)

= 40703

Therefore, the sum of the A.P. is 40703

 

(iii) (-5) + (-8) + (-11) + . . . + (-230)

Common difference of the A.P. (d) = a2 – a2

= -8 – (-5)

= -8 + 5

= -3

So here,

First term (a) = -5

Last term (l) = -230

Common difference (d) = -3

Let us take the number of terms as n.

an = a + (n – 1)d
So, the last term,

-230 = -5 + (n -1)(-3)

=> -230 = -5 -3n +3

=> -230 + 2 = -3n

=> -228 = -3n

=> n = 76

Substituting in the formula, we get,

\(S_{76} = \frac{76}{2} \left [ a + l \right ]\)

= 38[ (- 5) + (-230)]

= 38( -235 )

= -8930

Therefore, the sum of the A.P. is -8930

 

(iv) 1 + 3 + 5 + 7 + . . . + 199

Common difference of the A.P. (d)= a2 – a1

= 3 – 1

=2

So here,

First term (a) = 1

Last term (l) = 199

Common difference (d) = 2

Let us take the number of terms as n.

an = a + (n – 1)d

So, the last term,

199 = 1 + (n – 1)2

=> 199= 1 + 2n – 2

=> 199 + 1 = 2n

=> n = 100

Substituting in the formula, we get,

\(S_{100} = \frac{100}{2} \left [ a + l \right ]\)

= 50 [ 1 + 199 ]

= 50 (200)

= 10000

Therefore, the sum of the A.P. is 10000

 

(v) \(7 + 10\frac{1}{ 2} + 14 + . . . + 84\)

Common difference of the A.P. (d)= a2 – a1

= \(10\frac{1}{2} – 7\)

= \(\frac{21}{2} – 7\)

= \(\frac{21 – 14}{2}\)

= \(\frac{7}{2} \)

So here,

First term (a) = 7

Last term (l) = 184

Common difference (d) = \(\frac{7}{2} \)

Let us take the number of terms as n.

an = a + (n – 1)d

So, the last term,

\(84 = 7 + \left ( n – 1 \right )\frac{7}{2}\)

\(84 = 7 + \frac{7n}{2} – \frac{7}{2}\)

\(84 = \frac{14 – 7}{2} + \frac{7n}{2}\)

84 (2) = 7 + 7n

7n = 161

n = 23

Substituting in the formula,

\(S_{n} = \frac{23}{2}\left [ 2\left ( 7 \right ) + \left ( 23 – 1 \right )\frac{7}{2} \right ]\)

= \(\frac{23}{2}\left [ 14 + \left ( 22 \right )\frac{7}{2} \right ]\)

= \(\frac{23}{2}\left [ 14 + 77 \right ]\)

= \(\frac{23}{2}\left [91 \right ]\)

= \(\frac{2093}{2}\)

Therefore, the sum of the A.P. is \(\frac{2093}{2}\)

 

(vi) 34 + 32 + 30 + . . . + 10

Common difference of the A.P. (d) = a2 – a1

= 32 -34

= -2

So here,

First term (a) = 34

Last term (l) = 10

Common difference (d) = -2

Let us take the number of terms as n.

an = a  + (n – 1)d

So, the last term,

=> 10 = 34 + (n – 1)(-2)

=> 10 = 34 – 2n + 2

=> 10 = 36 – 2n

=> 10 – 36 = -2n

=> -2n = -26

=> n =13

Substituting in the formula, we get,

\(S_{n} = \frac{13}{2}\left [ a + l \right ]\)

= \(\frac{13}{2}\left [ 34 + 10 \right ]\)

= \(\frac{13}{2}\left [ 44 \right ]\)

= 12 (22)

= 286

Therefore, the sum of the A.P. is 286

 

(vii) 25 + 28 + 31 + . . . + 100

Common difference of the A.P. (d)= a2 – a1

= 28 – 25

=3

So here,

First term (a) = 25

Last term (l) = 100

Common difference (d) = 3

Let us take the number of terms as n.

an = a + (n – 1)d

100 = 25 + (n – 1)(3)

100 = 25 + 3n – 3

100 = 22 + 3n

100 – 22 = 3n

78 = 3n

n = 26

Substituting in the formula,

\(S_{n} = \frac{26}{2}\left [ a + l \right ]\)

= 13 [ 25 + 100 ]

= 13 (125)

= 1625

Therefore, the sum of the given A.P. is 1625

  

 14. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution :

The first term of the A.P (a) = 17

The last term of the A.P (l) = 350

The common difference of the A.P. = 9

Let the number of terms be n.

l = a + ( n – 1 )d

we get,

350 = 17 + ( n- 1 ) 9

350 =17 + 9n – 9

350 = 8 + 9n

350 – 8 = 9n

n=38

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Substituting the values, we get,

\(S_{n} = \frac{n}{2}\left [   a + l  \right ]\)

=> \(\frac{38}{2}\left ( 17 + 350 \right )\)

=> 19 X 367

=> 6973

Therefore, the number of terms is (n) 38 and the sum (Sn) is 6973 

 

15. The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.

Solution:

Let the first term = a

Let the common difference = d.

Given,

a3 = 7                                                                                                   . . . .(1)

a7  = 3a3 + 2                                                                                        . . . .(2)

From (1) in (2), we get,

a7 = 3(7) + 2

=21+2

= 23                                                                                                     . . . .(3)

We know that,

an = a +(n – 1)d

So, for the 3th term (n = 3),

a3 = a + (3 – 1)d

=> 7 = a + 2d

=> a =7 -2d                                                                                          . . . . (4)

Similarly, for the 7th term (n = 7),

a7 = a + (7 – I) d

24 = a + 6d                                                      (Using 3)

a = 24 – 6d                                                                                           . . . . (5)

Subtracting (4) from (5), we get,

a – a = (23 – 6d) – (7 – 2d)

=> 0 = 23 – 6d – 7 + 2d

=> 0 = 16 – 4d

=> 4d =16

=> d = 4

Substituting the value of d in (4) to find the value of a,

a =7 – 2(4)

=> a = 7 – 8

a = -1

So, for the given A.P, we have d = 4 and a = -1

So, to find the sum of first 20 terms of this A.P.,

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term for the given A.P.

d= common difference of the given A.P.

n= number of terms

So, using the formula for n= 20, we get,

\(S_{20} = \frac{20}{2}\left [ 2\left ( -1 \right ) + \left ( 20 – 1 \right )\left ( 4 \right ) \right ]\)

=(10)[-2 + (19) (4)]

= (10) [-2 + 76]

= (10) [74]

= 740

∴, the sum of first 20 terms for the given A.P. is S20 = 740

 

 

16. The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.

Solution:

Given,

The first term of the A.P (a) = 2

The last term of the A.P (I) = 50

Sum of all the terms S„ = 442

Let the common difference of the A.P. = d.

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Then,

\(442 = \left ( \frac{n}{2} \right )\left ( 2 + 50 \right )\)

=> \(442 = \left ( \frac{n}{2} \right )\left ( 52 \right )\)

=> 26 n = 442

=> n = 17

Now, to find the common difference of the A.P. we use the following formula,

l = a + (n – 1)d

We get,

50 = 2 + (17 – 1)d

=> 50 = 2 + 16d

=> 16d = 48

=> d = 3

Therefore, the common difference of the A.P. is d = 3

 

 

17. If 12th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

Solution:

Let the first term = a

Let the common difference = d

a12 = -13

S4 = 24

We know that,

an = a + (n – 1)d

For n = 12,

a12 = a + (12 – 1)d

-13 = a + 11d

a = -13 – 11d                                                                                        . . . .(1)

We also know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term for the given A.P.
d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 4, we get,

\(S_{4} = \frac{4}{2}\left [ 2\left ( a \right ) + \left ( 4 – 1 \right )d \right ]\)

=> 24 = (2) [ 2a + (3)(d) ]

=> 24 = 4a + 6d

=> 4a = 24 – 6d

=> \(a = 6 – \frac{6}{4}d\)                                                  . . . .(2)

Subtracting (1) from (2), we get.

=> \(a – a = \left (6 – \frac{6}{4}d \right ) – \left ( -13 – 11d \right )\)

=> \(0 = 6 – \frac{6}{4}d + 13d + 11d\)

=> \(0 = 19 + \frac{44d – 6d}{4}\)s

On further simplifying for d, we get,

=> \(0 = 19 + \frac{38}{4}d\)

=> -19 = \( \frac{19}{2}d\)

=> – 19 X 2 = 19 d

=> d = -2

Now, Substituting the value of d in (1),

a = -13 – 11 (-2)

a = -13 + 22

a = 9

Now, using the formula for the sum of n terms of an A.P., for n = 10

we obtain,

\(S_{10} = \frac{10}{2}\left [ 2\left ( 9 \right ) + \left ( 10 – 1 \right )\left ( -2 \right ) \right ]\)

= (5)[ 19 + (9)(-2) ]

= (5)(18 – 18)

= 0

Therefore, the sum of first 10 terms for the given A.P. is S10 = 0.

   

18. Find the sum of first 22 terms of an A.P. in which d = 22 and a22 = 149.

Solution:

Let the first term = a.

a22 = 149                                                                                  . . . .(1)

d = 22                                                                                     . . . .(2)

We know that,

an = a + ( n – 1) d

For n = 22,

a22 = a + (22 – 1) d

149 = a + (21) (22)                                                      (Using 1 and 2)

a = 149 – 462

a = – 313                                                                                              . . . .(3)

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

According to the question,

\(S_{22} = \frac{22}{2}\left [ 2\left ( -313 \right ) + \left ( 22 – 1 \right )\left ( 22 \right ) \right ]\)

= (11) [ -626 + 462 ]

= (11) [ -164 ]

= -1804

Therefore, The sum of first 22 terms for the given A.P. is S22 = -1804

 

 

19. In an A.P., if the first tern is 22, the common difference is -4 and the sum to n terms is 64, find n.

Solution:

a = 22

d = -4

S„ = 64

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula we get,

\(S_{n} = \frac{n}{2}\left [ 2\left ( 22 \right ) + \left ( n – 1 \right )\left ( -4 \right ) \right ]\)

\( 64 = \frac{n}{2}\left [ 2\left ( 22 \right ) + \left ( n – 1 \right )\left ( -4 \right ) \right ]\)

64(2) = n(48 – 4n)

128 = 48n – 4n2

The equation is of the form,

4n2 – 48n + 128 = 0

On taking 4 common, we get,

n2 – 12n + 32 = 0

Further, on solving the equation for n by splitting the middle term, we get,

n2 – 12n + 32 = 0

n2 – 8n – 4n + 32 = 0

n ( n – 8 ) – 4 ( n – 8 ) = 0

(n – 8) (n – 4) = 0

The equation implies,

n – 8 = 0

=> n = 8

Also,

n – 4 = 0

=> n = 4

Therefore, n = 4 or 8.

 

 

20. In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms ?

Solution:

Let the first term =a

Let the common difference = d

a5 = 30                                                                                                             . . . .(1)

a12 = 65                                                                                                           . . . .(2)

We know that,

an = a + (n – 1)d

For n = 5),

a5 = a + (5 – 1)d

30 = a + 4d                                                      (Using 1)

a = 30 – 4d                                                                                                      . . . .(3)

Similarly, for n = 12,

a12 = a + (12 – 1) d

65 = a + 11d                                                    (Using 2)

a = 65 – 11d                                                                                                     . . . .(4)
Subtracting (3) from (4), we get,

a – a = (65 – 11d) – (30 – 4d)

0 = 65 – 11d – 30 + 4d

0 = 35 – 7d

7d = 35

d = 5

Substituting the value of d in (4).

a = 30 – 4(5)

a = 30 – 20

a = 10

So, for the given A.P. d = 5 and a = 10

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where; a = first term of the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 20, we get

\(S_{20} = \frac{20}{2}\left [ 2\left ( 10 \right ) + \left ( 20 – 1 \right )\left ( 5 \right ) \right ]\)

= (10)[ 20 + (19)(5) ]

= (10)[20 + 95]

= (10)[115]

= 1150

Therefore, the sum of first 20 terms for the given A.P. is 1150

 

 

21. Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.

Solution:

a2 = 14                                                                                                 . . . . (1)

a3 = 18                                                                                                . . . . (2)

Also, we know,

an = a + (n – 1)d

For n = 2,

=> a2 = a + (2 – 1)d

=> 14 = a + d                                                   (Using 1)

=> a = 14 – d                                                                            . . . . (3)

Similarly, for n = 3,

=> a3 = a + (3 – 1)d

=> 18 = a + 2d                                                 (Using 2)

=> a = 18 – 2d                                                                          . . . . (4)
Subtracting (3) from (4),

we get, a – a = (18 – 2d) – (14 – d)

0 = 18 – 2d – 14 + d

0 = 4 – d

d = 4

Substituting the value of d in (4),

a = 14 – 4

a = 10

So, for the given A.P. d = 4 and a = 10

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where, a = the first term of the A.P.

d = common difference of the A.P.

n = number of terms

So, using the formula for n = 51, we get

\(S_{51} = \frac{51}{2}\left [ 2\left ( 10 \right ) + \left ( 51 – 1 \right )\left ( 4 \right ) \right ]\)

= \(\frac{51}{2}\left [ 20 + \left (40 \right )4 \right ]\)

= \(\frac{51}{2}\left [ 220  \right ]\)

= 51 (110)

= 5610

Therefore, the sum of the first 51 terms of the given A.P. is 5610

 

 

23. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution:

We know that,

Sum of n terms of an A.P.= \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

Where, a = the first term of the A.P.

d = common difference of the A.P.

n = number of terms

Also, nth term = an = a + (n – 1)

According to the question,

First term (a) = 5,

last term (an)= 45

and sum of n terms (Sn) = 400

Now,

an = a + (n – 1)d

45 = 5 + (n – 1)d

40 = nd – d

nd – d = 40                                                                               . . . . (1)

\(S_{n} = \frac{n}{2}\left ( 2(a) + (n – 1)d \right )\)

\(400 = \frac{n}{2}\left ( 2(5) + (n – 1)d \right )\)

800 = n ( 10 + nd – d )

800 = n (10 + 40)                                             from (1)

n  = 16                                                                                     . . . . (2)

On substituting (2) in (1), we get

nd – d = 40

16d – d = 40

15d = 40

d = \(\frac{8}{3}\)

Thus, common difference of the given A.P. is 83.

 

 

24. In an A.P. the first term is 8, nth term is 33 and the sum of first n term is 123. Find n and the d, the common difference.

Solution:

1st term of the A.P (a) = 8

nth term of the A.P (l)= 33

Sum of all the terms Sn = 123

Let the common difference of the A.P. be d.

So, let us first find the number of the terms (n) using the formula,

\(123 = \left ( \frac{ n }{ 2 } \right )\left ( 8 + 33 \right )\)

\(123 = \left ( \frac{ n }{ 2 } \right )\left ( 41 \right )\)

\(n = \frac{\left (123 \right )\left ( 2 \right )}{ 41 }\)

\(n = \frac{ 246 }{ 41 }\)

n = 6

To find the common difference,

l = a + ( n – 1 )d

33 = 8 + ( 6 – 1 )d

33 = 8 + 5d

5d = 25

d = 5

Therefore, the number of terms is n = 6 and the common difference of the A.P. is d = 5.

 

25. In an A.P. the first term is 22, nth term is -11 and the sum of first n term is 66. Find n and the d, the common difference.

Solution:

The first term of the A.P (a) = 22

The nth term of the A.P (l) = -11

Sum of all the terms S„ = 66

\(66 = \left ( \frac{ n }{ 2 } \right )\left [ 22 + \left ( -11 \right ) \right ]\)

\(66 = \left ( \frac{ n }{ 2 } \right )\left [ 22 – 11 \right ]\)

(66)(2) = n(11)

6 X 2 = n

n = 12

To find the common difference,

l = a + ( n – 1)d

we get,

-11 = 22 + ( 12 – 1 )d

-11 = 22 + 11d

11d = -33

d = -3

Therefore, the number of terms is n = 12 and the common difference d = -3

 

26. The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find the common difference.

Solution:

Let the first term=a

Let the common difference=b

We know that,

sum of first n terms,\(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

We also know that, nth term (an) = a + (n – 1)d

According to question,

first term (a) = 7

last term (an) = 49

and sum of n terms (Sn) = 420

Now,

an = a + (n – 1)d

=> 49 = 7 + (n – 1)d

=> 43 = nd – d

=> nd – d = 42                                                                                      . . . . .(1)

Sum of n terms, \(S_{n} = \frac{n}{2}\left ( 2 (7) + (n – 1)d \right )\)

=> 840 = n [14 + nd – d]

=> 840 = n [ 14 + 42 ]                                      [from (1)]

=> 840 = 54 n

=> n = 15                                                                                             . . . . (2)

on substituting (2) in (1), we get

nd – d = 42

=> 15d – d = 42

=> 14d = 42

=> d = 3

∴, the common difference of the given A.P. is 3.

 

28. The sum of first q terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1 : 2. Find the first and 15th term of the A.P.

Solution:

Let the first term=a

Let the common difference=b

We know that,

Sum of first n terms, \(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

Also, nth term = an = a + (n – 1) d

According to the question,

Sq = 162

and  a6 : a 13 = 1 : 2

Now,  2a6 = a13

=> 2 [a + (6 – 1d)] = a + (13 – 1)d

=> 2a + 10d = a + 12d

=> a = 2d                                                                                 . . . . (1)

Also, S = 162

=> \(S_{9} = \frac{9}{2}\left ( 2a + (9 – 1)d \right )\)

=> 162 = \( \frac{9}{2}\left ( 2a + 8d \right )\)

=> 162 X 2 = 9 [ 4d + 8d ]                    [from (1)]

=> 324 = 9 X 12d

=> d = 3

=> a =2d                                              [from (1)]

=> a = 6

Thus, the first term of the A.P. is 6

Now, a15 = a + 14d = 6 + 14 X 3 = 6 + 42

a 15 = 48

Therefore, 15th term of the A.P. is 48

 

29. If the 10th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth term.

Solution:

Let the first term=a

Let the common difference=b

We know that, sum of first n terms, \(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

and nth term is given by:

an = a + (n – 1)d

According to the question,

S10  = 120

=>        \(120 = \frac{10}{2}\left ( 2a + (10 – 1)d \right )\)

=>        120 = 5 (2a + 9d)

=>        24 = 2a + 9d                                                                            . . . .(1)

Also,

a10 = 21

=>        21 = a + (10 – 1)d

=>        21 = a + 9d                                                                              . . . . (2)

Subtracting (2) from (1), we get

24 – 21 = 2a + 9d – a – 9d

a = 3

Putting a = 3 in equation (2), we have

3 + 9d = 21

9d = 18

d = 2

first term = 3

common difference = 2

Therefore, the nth term can be calculated by:

an = a + (n – 1)d

= 3 + (n – 1) 2

= 3 + 2n -2

= 2n + 1

Therefore, the nth term of the A.P is (an) = 2n + 1

 

 

30. The sum of first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.

Solution:

Let the first term=a

Let the common difference=b

We know that, sum of first n terms, \(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

Given, sum of the first 7 terms of an A.P. is 63.

And sum of next 7 terms is 161.

Sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms

= 63 + 161 = 224

Now,    \(S_{7} = \frac{7}{2}\left ( 2a + (7 – 1)d \right )\)

=> 63 (2) = 7 ( 2a + 6d )

=> 9 X 2 = 2a + 6d

=> 2a + 6d = 18                                                           . . . . (1)

Also,    \(S_{14} = \frac{14}{2}\left ( 2a + (14 – 1)d \right )\)

=> 224 = 7 ( 2a + 13d )

=> 32 = 2a + 13d                                                         . . . . (2)

On subtracting (1) from (2),

=> 13d – 6d = 32 – 18

=> 7d = 14

=> d = 2

From (1)

2a + 6 (2) = 18

2a = 18 – 12

a = 3

Also, nth term = an = a + (n – 1)d

=> a28 = a + (28 – 1)d

= 3 + 27 (2)

= 3 + 54

= 57

Thus, the 28th term is 57.

 

 

31. The sum of first seven terms of an A.P. is 182. If its 4th and 17th terms are in ratio 1 : 5, find the A.P.

Solution:

Let the first term=a

Let the common difference=b

S17 = 182

We know that,

Sum of first term, \(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

According to the question,

\(S_{7} = \frac{7}{2}\left ( 2a + (7 – 1)d \right )\)

182 X 2 = 7 ( 2a + 6d )

364 = 14a + 42d

26 = a + 3d

a = 26 – 3d                                                                                           . . . (1)

4th term:17th term are in a ratio 0f 1 : 5

5 (a4) = 1 (a17)

5 ( a + 3d) = 1 ( a +16d)

5a + 15d = a + 16d

4a = d                                                                                             . . . . (2)

On substituting (2) in (1), we get

4 ( 26 – 3d ) = d

104 – 12d = d

104 = 13d

d = 8

from (2), we get

=> 4a = d

=> 4a = 8

=> a = 2

Thus we get, first term a = 2 and the common difference d = 8.

The required A.P. is 2, 10, 18, 26, . . .

 

33. In an A.P. the sum of first ten terms is -150 and the sum of its next 10 term is -550. Find the A.P.

Solution:

Let the first term=a

Let the common difference=b

So, let us first find S10.

For the sum of first 10 terms of this A.P,

First term = a

Last term = a10

So, we know,

an = a +(n – 1)d

For the 10th term (n = 10),

an = a + (10 – 1)d

= a + 9d

We know that,

Sum of the n terms, \(S_{n} = \left ( \frac{ n }{ 2 } \right )\left ( a + l \right )\)

Where, a = the first term

l = the last term

According to the question,

\(S_{10} = \left ( \frac{ 10 }{ 2 } \right )\left ( a + a + 9d \right )\)

-150 = 5 (2a + 9d)

-150 = 10a + 45 d

\(a = \frac{ 150 – 45d }{ 10 }\)                                           . . . . (1)

First term = a11

Last term = a20

For n = 11,

a11 = a+ (11 – 1) d

= a + 10d

For n = 20,

a20 = a + (20 – 1) d

= a + 19d

So, for the given AP,

\(S_{10} = \left ( \frac{10}{2} \right )\left ( a + 10d + a + 19d \right )\)

-550 = 5 (2a + 29d)

-550 = 10a + 145d

\(a = \frac{-550 – 145 d}{10}\)                                           . . . . (2)

Now subtracting (1) from (2),

a – a = \(\left (\frac{-550 – 145 d}{10} \right ) – \left ( \frac{ -150 – 45d }{ 10 } \right )\)

0 = -550 – 145d + 150 + 45d

0 = -400 – 100d

100d = – 400

d = -4

Substituting the value of d in (1)

\(a = \left (\frac{-150 – 45 \left ( -4 \right )}{10} \right )\)

\(a = \left (\frac{-150 + 180}{10} \right )\)

= 3

So, the A.P. is 3, -1, -5, -9,. . .  with a = 3, d= -4

 

35. In an A.P. , the first term is 2, the last term is 29 and the sum of the terms is 155, find the common difference of the A.P.

Solution:

The first term of the A.P (a) = 2

The last term of the AP (l) = 29

Sum of all the terms (Sn) = 155

Let the common difference of the A.P. =d.

\(155 = \frac{n}{2}\left ( 2 + 29 \right )\)

155 (2) = n (31)

31n =310

n = 10

To find the common difference, d,

l = a + (n – 1) d

We get,

29 = 2 + (10 – 1)d

29 = 2 + (9)d

29-2 = 9d

9d = 27

d = 3

Therefore, the common difference of the A.P. is d = 3

 

37. Find the number of terms of the A.P. -12, -9, -6, . . . , 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.

Solution:

First term, a1 = -12

Common difference, d = a2 – a1 = -9 – ( – 12)

= – 9 + 12 = 3

nth term =        an = a + (n – 1)d

=> 21 = -12 + (n – 1)3

=> 21 = -12 + 3n – 3

=> 21 = 3n – 15

=> 36 = 3n

=>  n = 12

1 added to each of the 12 terms ⇒ the sum will increase by 12.

We know that,

Sum of the n terms, \(S_{n} = \left ( \frac{ n }{ 2 } \right )\left ( a + l \right )[/latex

=> S12 + 12  \(= \frac{12}{2}\left [ a + l \right ] + 12\)

= 6 [-12 + 21 ] + 12

= 6 X 9 + 12

= 66

∴, the sum after adding 1 to each of the term we get 66

 

 

38. The sum of first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.

Solution:

let the first term = a

let the common difference =d.

nth term, an = Sn  – Sn-1

Sn = 3n2 + 6n

an = [3n2 + 6n] – [3(n – 1)2 + 6 (n – 1)]

= [3n2 + 6n] – [ 3 (n2 + 12 – 6n) + 6n – 6 ]

= 3n2 + 6n – 3n2 – 3 + 6n – 6n + 6

= 6n + 3

Therefore, the nth term of this A.P. is 6n + 3

 

39. The sum of n terms of an A.P. is 5n – n2. Find the nth term of this A.P.

Solution:

let the first term = a

let the common difference =d.

We know that,

Sum of first n terms is,\(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

Given, sum of the first n terms of an A.P. is 5n – n2.

First term = a = S1 = 5 (1) – (1)2 = 4.

Sum of first two terms = S2= 5 (2) – (2)2 = 6.

Second term = S2 – S1 = 6 – 4 = 2.

Common difference = d = Second term – First term

= 2 – 4 = -2

Also, nth term = an = a + (n – 1) d

=> an = 4 + (n – 1)(-2)

=> an = 4 – 2n + 2

=> an = 6 – 2n

Thus, nth term of this A.P. is 6 – 2n.

 

41. The sum of first n terms of an A.P. is 3n2+ 4n. Find the 25th term of this A.P.

Solution:

Sn = 5n2 + 3n

an = Sn – Sn-1

For n = 25

a25 = S25 – S24

= [3 (25)2 + 4 (25)] – [ 3 (24)2 + 4 (24) ]

= (3 X 625 + 100) – ( 3 X 576 + 96 )

= 1975 – 1824

= 151

Therefore, its 25th term is 151

 

42. The sum of first n terms of an A.P. is 5n2 + 3n. If its mth term is 168, find the value of m. Also find the 20th term of this A.P.

Solution:

Sum of the A.P. as  Sn = 5n2 + 3n.

mth term is a= 168

Let first term =a,

Let common difference = d

an = Sn – Sn-1

an = (5n2 + 3n) – [ 5(n – 1)2 + 3 (n – 1)]

= 5n2 + 3n – [ 5( n2 + 1 – 2n) + 3n – 3 ]

= 5n2 + 3n – 5n2 -5 + 10n – 3n + 3

= 10n – 2

am = 168

Putting m in place of n , we get

=> am = 10m – 2

= > 168 = 10m – 2

=> 10m = 170

=> m = 17

a20 = S20 – S19

= [5 (20)2 + 3 (20)] – [5 (19)2 + 3 (19) ]

= [2000 + 60] – [1805 + 57]

= 2060 – 1862

= 198

Therefore, in the given A.P. m = 17 and the 20th term is a20  = 198

 

45. If the sum of first n terms of an A.P. is 4n – n2 , what is the first term? What is the sum of first two terms? What is the second term? Similarly find the third, the tenth and the n­th term.

Solution:

The sum of n terms of an A.P. Sn = 4n – n2

By substituting n = 1,

Sn = 4n – n2

= 4(1)—12

= 4 – 1

= 3

Sum of first two terms,

S2 = 4(2) – (2)2

= 8 – 4

= 4

an = Sn – Sn-1

a2 = S2 – S1

= 4 – 3

= 1

a3 = S3 – S2

=[4 (3) – (3)2] – [4(2) – (2)2]

= (12 – 9) – (8 – 4)

= 3 – 4

= -1

For n=10

a10   = S10 – S9

= [44(10) – (10)2]  – [4(9) – (9)2]

=(40 – 100) – (36 – 81)

= – 60 + 45

= -15

So, for the nth term,

an = Sn – S­n-1

=[4(n) – (12)2] – [4(n – 1) – (n – 1)2]

= (4n –  n2) – (4n – 4 – n2 – 1 + 2n)

= 4n – n2 – 4n + 4 + n2 + 1 – 2n

= 5 – 2n

Therefore, a = 3, S2 = 4, a2  = 1, a3 = -1, a10 = -15

 

 

46. If the sum of first n terms of an A.P. is \(\frac{1}{2}\left ( 3n^{2} + 7n \right )\) , then find its nth term. Hence write the 20th term.

Solution:

Let first term=a

Let common difference=d

We know that, sum of first n terms,\(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

Given, sum of the first n terms of an A.P. is:

\(\frac{1}{2}\left ( 3n^{2} + 7n \right )\)

Therefore, first term (a) = \(S­_{1} = \frac{1}{2}\left ( 3(1)^{2} + 7(1) \right )\)

= \(\frac{1}{2}\left ( 3 X 1 + 7 \right )\)

= \(\frac{1}{2}\left ( 10 \right )\)

= 5

Sum of first 2terms = \(S­_{2} = \frac{1}{2}\left ( 3(2)^{2} + 7(2) \right )\)

= \(\frac{1}{2}\left ( 3 X 4 + 14 \right )\)

= \(\frac{1}{2}\left ( 26 \right )\)

= 13

Now, second term = S2 – S1

= 13 – 5 = 8

Common difference, d = second term – first term

= 8 – 5 = 3

Also, nth term of the A.P. is : a + (n – 1)d

= 5 + (n – 1)3

= 5 + 3n – 3

= 3n + 2

Thus, nth term of this A.P. is 3n + 2.

To find the 20th term, n = 2o, we get

a20 = 3 (20) + 2

= 60 + 2

= 62

Thus, 20th term of this A.P. is 62.

 

47. In an A.P. the sum of first n terms is \(\frac{3n^{2}}{2} + \frac{13}{2}n\). Find its 25th term.

Solution:

We know that,

Sum of first n terms,\(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

\(S_{n} = \frac{3n^{2}}{2} + \frac{13}{2}n\)

nth term of an A.P, an = Sn – Sn-1

So, a25 = S25 – S24                                                                                                         . . . . (1)

Substituting in the formula,

\(S_{n} = \frac{3\left ( 25 \right )^{2}}{2} + \frac{13}{2}\left ( 25 \right )\)

= \(\frac{3\left ( 625 \right )}{2} + \frac{325}{2}\)

= \(\frac{2200}{2}\)

= 1100

Similarly,

\(S_{n} = \frac{3\left ( 24 \right )^{2}}{2} + \frac{13}{2}\left ( 24 \right )\)

= \(\frac{3\left ( 576 \right )}{2} + \frac{312}{2}\)

= \(\frac{2040}{2}\)

= 1020

Now, using the above values in (1),

a21 = S25 – S24

= 1100 – 1020

= 80

Therefore, a25 = 80

 

 

48. Find the sum of all natural numbers between 1 and 100, which are divisible by 3.

Solution:

First term (a) = 3

Last term (l) = 99

Common difference (d) = 3

Let the number of terms =n.

an = a + ( n – 1 )d

99 = 3 + (n – 1)3

99 = 3 + 3n – 3

99 = 3n

n = 33

We know that,

Sum of first n terms,\(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

i.e. \(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

we get,

=> \(S_{33} = \frac{33}{2}\left [ 2\left ( 3 \right ) + \left ( 33 – 1 \right )3 \right ]\)

= \(S_{33} = \frac{33}{2}\left [ 6 + \left ( 32 \right )3 \right ]\)

= \(S_{33} = \frac{33}{2}\left [ 6 + 96 \right ]\)

= \(S_{33} = \frac{33\left ( 102 \right )}{2}\)

= 33 ( 51 )

= 1683

Therefore, the sum of all the multiples of 3 lying between 1 and 100 is Sn = 1683

 

50. Find the sum of all odd numbers between (i) 0 and 50 (ii) 100 and 200.

Solution:

(i)

First term (a) = 1

Last term (0 = 49

Common difference (d)= 2

According to the question,

an = a + (n -1)d

49 = 1 + (n – 1)d

49 = 1 + 2n – 2

49 = 2n – 2

49 + 1 = 2n

50 = 2n

n = 25

We know that,

Sum of first n terms,\(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

for n = 25,

\(S_{25} = \frac{25}{2}\left [ 2(1) + \left ( 25 – 1 \right )2 \right ]\)

= \(\frac{25}{2}\left [ 2 + 24 \times 2 \right ]\)

= 25 X 25

= 625

Therefore, the sum of all the odd numbers lying between 0 and 50 is 625.

 

(ii)

First term (a) = 101

Last term (an) = 199

Common difference (d)= 2

an = a + ( n – 1)d

199 = 101 + (n – 1)2

199 = 101 + 2n – 2

199 = 99 + 2n

199 – 99 = 2n

100 = 2n

n = 50

We know that,

Sum of first n terms,\(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

\(S_{n} = \frac{n}{2}\left [ 2\left ( a \right ) + \left ( n – 1 \right )d \right ]\)

For n = 50,

\(S_{50} = \frac{50}{2}\left [ 2\left ( 101 \right ) + \left ( 50 – 1 \right )2 \right ]\)

= 25 [ 202 + (49) 2 ]

= 25( 202 + 98 )

= 25 (300)

= 7500

Therefore, the sum of all the odd numbers lying between 100 and 200 is 7500

 

 

52. Find the sum of all integers between 84 and 719, which are multiples of 5.

Solution:

First term (a) = 85

Last term (l) = 715

Common difference (d) = 5

Let us take the number of terms as n.

a = a +(n -1)d

715 = 85 + (n – 1)5

715 = 85 + 5n – 5

715 = 80 + 5n

715 – 80 = 5n

635 = 5n

n = 127

We know that,

Sum of first n terms,\(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

\(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

For n = 127,

\(S_{127} = \frac{127}{2}\left [ 2(85) + \left ( 127 – 1 \right )5 \right ]\)

= \(\frac{127}{2}\left [ 170 + 630 \right ]\)

= \(\frac{127 \left ( 800 \right ) }{2}\)

= 50800

Therefore, the sum of all the multiples of 5 lying between 84 and 719 is 50800.

 

  

53. Find the sum of all integer between 50 and 500, which are divisible by 7.

Solution:

First term (a) = 56

Last term (l) = 497

Common difference (d) = 7

an = a  + (n – 1)d

497 = 56 + ( n – 1 )7

497 = 56 + 7n -7

497 = 49 + 7n

497  – 49 = 7n

448 = 7n

n = 64

We know that,

Sum of first n terms,\(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

\(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

for n = 64,

\(S_{64} = \frac{64}{2}\left [ 2\left ( 56 \right ) + \left ( 64 – 1 \right )7 \right ]\)

= 32 [ 112 + (63)7]

= 32 [112 + 441]

=32 (553)

= 17696

Therefore, the sum of all the multiples of 7 lying between 50 and 500 is 17696

 

54. Find the sum of all even integers between 101 and 999.

Solution:

First term (a) = 102

Last term (l) = 998

Common difference (d) = 2

an = a  + (n – 1)d

998 = 102 + (n – 1)2

998 = 102 + 2n – 2

998 = 100 + 2n

998 – 100 = 2n

898 = 2n

n = 449

We know that,

Sum of first n terms,\(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

\(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

For n = 449,

\(S_{449} = \frac{449}{2}\left [ 2\left ( 102 \right ) + \left ( 449 – 1 \right )2 \right ]\)

= \(\frac{449}{2}\left [ 204 + \left ( 448 \right )2 \right ]\)

= \(\frac{449}{2}\left [ 204 + 896 \right ]\)

= \(\frac{449}{2}\left [ 1100 \right ]\)

= 449 ( 550 )

= 246950

Therefore, the sum of all even numbers lying between 101 and 999 is  246950

 

 

55.
(i) Find the sum of all integers between 100 and 550, which are divisible by 9.

Solution:

First term (a) = 108

Last term (l) = 549

Common difference (d) = 9

an = a + (n – 1)d

549 = 108 + (n – 1)d

549 = 108 + 9n – 9

549 = 99 + 9n

549 – 99 = 9n

9n = 450

n = 50

We know that,

Sum of first n terms,\(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

\(S_{n} = \frac{n}{2}\left [ 2a + \left ( n – 1 \right )d \right ]\)

\(S_{n} = \frac{50}{2}\left [ 2\left ( 108  \right ) + \left ( 50 – 1 \right )9 \right ]\)

= 25 [ 216 + (49)9 ]

= 25 (216 + 441)

= 25 (657)

= 16425

Therefore, the sum of all the multiples of 9 lying between 100 and 550 is 16425

 

56. Let there be an A.P. with first term ‘a’, common difference ‘d’. If an denotes its nth term and Sn the sum of first n terms, find.

(i)         n and Sn, if a = 5 , d = 3 , and an = 50.

(ii)        n and a, if an = 4 ,  d = 2 and Sn = -14.

(iii)       d, if a = 3, n = 8 and Sn = 192.

(iv)       a, if an = 28, Sn = 144 and n = 9.

(v)        n and d, if a = 8, an = 62 and S­n­ ­= 120.

(vi)       n and an , if a = 2, d = 8 and Sn = 90. 

Solution:

(i)

First term (a) = 5

Last term (an) = 50

Common difference (d) = 3

Using the formula, an = a + (n – 1) d

50 = 5 + (n – 1) 3

50 = 5 + 3n – 3

50 = 2 + 3n

3n = 50 – 2

3n = 48

n =16

We know that,

Sum of first n terms, \(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

Where,

a = the first term

l = the last term

According to the question,

\(S_{16} = \left ( \frac{16}{2} \right )\left ( 5 + 50 \right )\)

= 8 (55)

= 440

∴, for the given A.P. we have, n = 16 and S16 = 440

 

 (ii)

Last term (l) = 4

Common difference (d) = 2

Sum of n terms (Sn) = -14

Using the formula, an = a + (n – 1) d

4 = a + ( n – 1 ) 2

4=a+2n-2

4 + 2 = a + 2n

n = \(\frac{6 – a}{2}\)                                                    . . . . (1)

We know that,

Sum of first n terms, \(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

Where, a = the first term

l = the last term

According to the question,

\(-14 = \frac{n}{2} \left ( a + 4 \right )\)

14 (2) = n (a + 4)

\(n = \frac{-28}{a + 4}\)                                                            . . . . (2)

Equating (1) and (2), we get,

\(\frac{6 – a}{2} = \frac{-28}{a + 4}\)

(6 – a)(a + 4) = -28(2)

6a – a2 + 24 – 4a = -56

-a2 + 2a + 24 + 56 = 0

-a 2 + 2a + 80 = 0

a2 – 2a – 80 = 0

a2 – 2a – 80 = 0

a2 – 10a + 8a – 80 = 0

a(a – 10) + 8(a – 10) = 0

(a – 10)(a + 8) = 0

So, we get,

a – 10 = 0

a = 10

or,

a + 8 = 10

a = -8

Substituting, a = 10 in (1)

\(n = \frac{ 6 – 10 }{ 2 }\)

\(n = \frac{ -4 }{ 2 }\)

n = -2

Here, we get n as negative, which is not possible to occur.

So, we take a = -8

\(n = \frac{ 6 – \left ( -8 \right ) }{ 2 }\)

\(n = \frac{ 6 + 8 }{ 2 }\)

\(n = \frac{ 14 }{ 2 }\)

n = 7

∴, for the given A.P. n = 7 and a = -8

  

(iii)

First term ( a ) = 3

Sum of n terms (Sn) = 192

Number of terms (n) = 8

Using the formula, an = a + (n – 1) d

We know that,

Sum of first n terms, \(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

Where;

a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 8, we get,

\(S_{8} = \left ( \frac{8}{2} \right )\left [ 2 (3) + \left ( 8 – 1 \right ) d \right ]\)

192 = 4 [ 6 + 7d ]

192 = 24 + 28d

28d = 192 – 24

28 d = 168

d = 6

∴, the common difference of the given A.P. is d = 6

  

(iv)

Last term ( a9) = 28

Sum of n terms (Sn)= 144

Number of terms (n) = 9

Now,

a9 = a + 8d

28 = a +8d                                                                                           . . . . (1)

We know that,

Sum of first n terms, \(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 9, we get,

\(S_{9} = \left ( \frac{9}{2} \right )\left [ 2a + \left ( 9 – 1 \right ) d \right ]\)

144 (2) = 9 [2a + 8d]

288 = 18a + 72d                                                                                  . . . . (2)

Multiplying (1) by 9, we get

9a +72d = 252                                                                                     . . . .  (3)

Further, subtracting (3) from (2), we get

9 a = 36

a = 4

∴, the first term of the given A.P. is a = 4

 

(v)

First term (a ) = 8

Last term ( an ) = 62

Sum of n terms (Sn) = 210

We know that,

Sum of first n terms, \(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

Where, a = the first term

l = the last term

According to the question,

\(210 = \frac{n}{2} \left [ 8 + 62 \right ]\)

210 (2) = n (70)

\(n = \frac{420}{70}\)

n = 7

Using the formula, an = a + (n – 1) d

62= 8 + (6 – 1)d

5d = 54

\(d = \frac{54}{5}\)

∴, for the given A.P. n = 6 and \(d = \frac{54}{5}\)

 

(vi)

First term (a) = 2

Sum of first nth terms (Sn) = 90

Common difference (d) = 8

We know that,

Sum of first n terms, \(S_{n} = \frac{n}{2}\left ( 2a + (n – 1)d \right )\)

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for d = 8, we get,

\(S_{n} = \left ( \frac{n}{2} \right )\left [ 2 (2) + \left ( n – 1 \right ) 8 \right ]\)

\(90 = \left ( \frac{n}{2} \right )\left [ 4 + 8n – 8 \right ]\)

90 (2) = n [ 8n – 4 ]

180 = 8n2 – 4n

8n2 -4n -180 = 0

2n2 – n – 45 =0

2n2 – 10n + 9n – 45 = 0

2n(n – 5) + 9( n – 5 ) = 0

(2n – 9)(n – 5) = 0

From the equation,

2n + 9 = 0

\(n = \frac{-9}{2}\)

n – 5 = 0

n = 5

Since, n cannot be a fraction.

Thus, n = 5

Using the formula, an = a + (n – 1) d

an = 2 + (5 – 1) 8

an = 2 + 4 (8)

an = 2 + 32

an = 34

∴, for the given A.P., n = 5 and an = 34

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