RS Aggarwal Class 10 Solutions Arithmetic Progressions

An arithmetic sequence which is a simple progression of numbers, where the difference between the consecutive terms is constant is also called Arithmetic progressions. The sum of all the numbers which is a part of the arithmetic progression is titled as an arithmetic series. When a progression has a finite amount of numbers, they are called finite arithmetic progressions whereas a sequence with an infinite set of numbers is called as an infinite set of progressions.

Check out RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions below:

Q.1: Show that each of the progressions given below is an AP. 1: 1: Find the first term, common difference and next term of each.

(i) 3, 9, 15, 21…..

Solution:

The given progression is 3, 9, 15, 21…..

Clearly (9 – 3) = (15 – 9) = (21 – 15) = 6 which is constant

Thus, each term differs from its preceding term by 6.

So, the given progression is an AP.

Next term of the AP = 21 + 6 = 28

Its first term = 3, common difference = 6 and the next term is 28.

(ii) 16, 11, 6, 1, -4….

The given progression is 16, 11, 6, 1, -4….

Clearly (11 – 16) = (1 – 6) = (-4 – 1) = – 5 which is constant.

Thus, each term differs from its preceding term by – 5.

So the given progression is an AP.

Next term of the AP = -4 – (-5) = -9

Its first term = 16 , common difference = – 5 and the next term is -9

(iii) \(-1, -\frac{5}{6}, -\frac{2}{3}, -\frac{1}{2},……\)

The given progression is \(-1, -\frac{5}{6}, -\frac{2}{3}, -\frac{1}{2},……\)

Clearly \(-\frac{5}{6} – (-1) = -\frac{1}{2} = \frac{-2}{3} = \frac{1}{6}\)

Thus, each term differs from its preceding term by \(\frac{1}{6}\). So, the given progression is an AP.

First term = -1

Common difference = \(\frac{1}{6}\)

Next term of the AP = \(\frac{-1}{2} + \frac{1}{6} = \frac{-2}{6} = \frac{-1}{3}\)

(iv) \(\sqrt{2}, \sqrt{8}, \sqrt{18} , \sqrt{32} , …….\)

The given progression \(\sqrt{2}, \sqrt{8}, \sqrt{18} , \sqrt{32} , …….\)

This sequence can be re-written as \(\sqrt{2}, 2\sqrt{2}, 3\sqrt{2} , 4\sqrt{2} , …….\)

d = \(2\sqrt{2} – 2 = \sqrt{2}\)

Thus, each term differs from its preceding term by \(\sqrt{2}\). So, the given progression is an AP.

First term = \(\sqrt{2}\)

Common difference = \(\sqrt{2}\)

Next term of an AP = \(4 \sqrt{2} + \sqrt{2} = 5 \sqrt{2}\)

\(5 \sqrt{2} = \sqrt{50}\)

(v) \(\sqrt{20}, \sqrt{15}, \sqrt{80} , \sqrt{125} , …….\)

The given progression is \(\sqrt{20}, \sqrt{15}, \sqrt{80} , \sqrt{125} , …….\)

This sequence can be re-written as \(2 \sqrt{5}, 3 \sqrt{5}, 4 \sqrt{5} , 5 \sqrt{5} , …….\)

Clearly d = \(3 \sqrt{5} – 2 \sqrt{5} = \sqrt{5}\)

Thus, each term differs from its preceding term by \(\sqrt{5}\). So, the given progression is an AP.

First term = \(2 \sqrt{5} = \sqrt{20}\)

Common difference = \(\sqrt{5}\)

Next term of the AP = \(5 \sqrt{5} + \sqrt{5} = 6 \sqrt{5} = \sqrt{180}\)

Q.2: Find:

(i) The 20th term of the AP 9, 13, 17, 21,………

The given AP is 9, 13, 17, 21,………

First term, a = 9

Common difference, d = 13 – 9 = 4

nth term of the AP, \(a_{n} = a + (n – 1)d = 9 + (n – 1) \times 4\)

Therefore, 20th term of the AP, \(a_{20} = 9 + (20 – 1) \times 4 = 9 + 76 = 85\)

(ii) the 35th term of the AP 20, 17, 14, 11, ……..

The given AP is 20, 17, 14, 11, ……..

First term, a = 20

Common difference, d = 17 – 20 = -3

nth term of the AP, \(a_{n} = a + (n – 1)d = 20 + (n – 1) \times (-3)\)

Therefore, 35th term of the AP, \(a_{35} = 20 + (35 – 1) \times (-3) = 20 – 102 = – 82\)

(iii) the 18th term of the AP \(\sqrt{2}, \sqrt{18}, \sqrt{50},\sqrt{98}, ………\)

The given AP is \(\sqrt{2}, \sqrt{18}, \sqrt{50},\sqrt{98}, ………\)

This can be re-written as \(\sqrt{2}, 3 \sqrt{2},5\sqrt{2}, 7 \sqrt{2} ……….\)

First term, a = \(\sqrt{2} \)

Common difference, d = \(3 \sqrt{2} – \sqrt{2} = 2 \sqrt{2}\)

nth term of the AP, \(a_{n} = a + (n – 1)d = \sqrt{2} + (n – 1) \times 2 \sqrt{2} \)

Therefore, 18th term of the AP, \(a_{18} = \sqrt{2} + (18 – 1) \times 2 \sqrt{2} = \sqrt{2} + 34 \sqrt{2} = 35 \sqrt{2} = \sqrt{2450}\)

(iv) the 9th term of the AP \(\frac{3}{4}, \frac{5}{4} , \frac{7}{4}, \frac{9}{4}, ……\)

The given AP is \(\frac{3}{4}, \frac{5}{4} , \frac{7}{4}, \frac{9}{4}, ……\)

First term, a = \(\frac{3}{4}\)

Common difference, d= \( \frac{5}{4} – \frac{3}{4} =  \frac{2}{4} = \frac{1}{2} \)

nth term of the AP, \(a_{n} = a + (n – 1) d = \frac{3}{4} + (n – 1) \times \frac{1}{2} \)

Therefore, 9th term of the AP, \(a_{9} = \frac{3}{4} + (9 – 1) \times \frac{1}{2} = \frac{3}{4} + 4 = \frac{19}{4} \)

(v) the 15th term of the AP  -40, -15, 10, 35, ……

The given AP is  -40, -15, 10, 35, ……

First term, a = -40

Common difference, d = -15 – (-40) = 25

nth term of the AP, \(a_{n} = a + (n – 1)d = -40 + (n – 1) \times 25 \)

Therefore, 15th term of the AP, \(a_{15} = -40 + (15 – 1) \times 25 = -40 + 350 = 310 \)

Q.3: Find the 37th term of the AP 6, \(7 \frac{3}{4}, 9 \frac{1}{2}, 11 \frac{1}{4}, ………..\)

Sol:

The given AP is 6, \(7 \frac{3}{4}, 9 \frac{1}{2}, 11 \frac{1}{4}, ………..\)

First term, a = 6

Common difference, d = \(7 \frac{3}{4} – 6 = \frac{7}{4}\)

nth term of the AP, \(a_{n} = a + (n – 1)d = 6 + (n – 1) \times \frac{7}{4} \)

Therefore, 37th term of the AP, \(a_{37} = 6 + (37 – 1) \times \frac{7}{4} = 6 + 63 = 69 \)

Therefore, 37th term = 69

Q.4: Find the 25th term of the AP 5, \(4 \frac{1}{2}, 4, 3 \frac{1}{2}, 3 , ………..\)

Sol:

The given AP is 5, \(4 \frac{1}{2}, 4, 3 \frac{1}{2}, 3 , ………..\)

First term, a = 5

Common difference, d = \(4 \frac{1}{2} – 5 = \frac{-1}{2}\)

nth term of the AP, \(a_{n} = a + (n – 1)d = 5 + (n – 1) \times \frac{-1}{2} \)

Therefore, 25th term of the AP, \( a_{25} = 5 + (25 – 1) \times \frac{-1}{2} = 5 – 12 = -7 \)

Therefore, 25th term = -7

Q.5: Find the nth term of each of the following APs:

(i) 5, 11, 17, 23,……..  

(ii) 16, 9, 2, -5, ………

Sol:

(i) The given AP is 5, 11, 17, 23,……..

First term, a = 5

Common difference, d = 11 – 5 = 6

nth term of the AP, \(a_{n} = a + (n – 1)d = 5 + (n – 1) \times 6 \)

= 5 + 6n – 6

= 6n – 1

(ii) The given AP is 16, 9, 2, -5, …..

First term, a = 16

Common difference, d = 9 – 16 = -7

nth term of the AP, \(a_{n} = a + (n – 1)d = 16 + (n – 1) \times (-7) \)

= 16 – 7n + 7

= 23 – 7n

Q.6: If the nth term of the progression is (4b – 10) show that it is an AP. Find its:

(i) first term

(ii) common difference

(iii) 16th term

Sol:

Given : \(T_{n} = 4n – 10\)

\(T_{1} = 4 \times 1  – 10 = -6\)

\(T_{2} = 4 \times 2  – 10 = -2\)

\(T_{3} = 4 \times 3  – 10 = 2\)

\(T_{4} = 4 \times 4  – 10 = 6\)

Clearly d = -2 – (-6) = 4

So, the terms -6, -2, 2, 6, …….. forms an AP.

Thus, we have:

(i) First term = -6

(ii) Common difference = 4

(iii) \(T_{16} = -6 + (16 – 1) \times 4 = 54\)

Q.7: How many terms are there in an AP 6, 10, 14, 18, ………, 174 ?

Sol:

In the given AP, we have a = 6 and d = (10 – 6) = 4

Suppose there are n terms in the given AP, then

\( T_ {n} \)    = 174

\( \Rightarrow \) a + (n – 1) d = 174

\( \Rightarrow \) 6 + (n – 1) 4 = 174

\( \Rightarrow \) 6 + 4n – 4 = 174

\( \Rightarrow \) 2 + 4n = 174

\( \Rightarrow \) n = (\( \frac {172} {4} \))

\( \Rightarrow \) 43

Hence there are 43 terms in the given AP.

Q.8: How many terms are there in an AP 41, 38, 35, ………, 8?

Sol:

In the given AP we have a = 41 and d = 38 – 41 = – 3

Suppose there are n terms in AP, then

\( T_ {n} \)   = 8

\( \Rightarrow \) a + (n – 1) d = 8

\( \Rightarrow \) 41 + (n – 1) (-3) = 8

\( \Rightarrow \) 41 – 3n + 3 = 8

\( \Rightarrow \) -3n = – 36

\( \Rightarrow \) n = 12

Hence there are 12 terms in the given AP.

Q.9: How many terms are there in the AP \(18, 15\frac{1}{2}, 13, …… ,-47\)

Sol:

The given AP is \(18, 15\frac{1}{2}, 13, …… ,-47\)

First term, a = 18

Common difference, d = \(15\frac{1}{2} – 18 = -\frac{5}{2}\)

Suppose there are n terms in the given AP.

Then, \(a_{n} = -47\)

\(\Rightarrow 18 + (n – 1) \times \left ( -\frac{5}{2} \right ) = – 47\)

\(\Rightarrow \left ( -\frac{5}{2} \right ) (n – 1)= – 65\)

\(\Rightarrow n – 1 = 65 \times (- \frac{2}{5}) = 26\)

\(\Rightarrow n = 27\)

Hence, there are 27 items in the given AP.

Q.10: Which term of the AP 3, 8, 13, 18, ………, is 88?

Sol:

In the given AP, we have a = 3 and d = 8 – 3 = 5

Suppose there are n terms in given AP, then

\(T_{n}\)       = a + (n-1) d = 88

\(\Rightarrow\)  3 + (n-1) 5 = 88

\(\Rightarrow\)  3 + 5n – 5 = 88

\(\Rightarrow\)  5n = 90

\(\Rightarrow\)   n = 18

Hence, the \(18^{th}\)  term of given AP is 88

Q.11: Which term of the AP 72, 68, 64, 60, …….., is 0?

Sol:

In the given AP, we have a = 72 and d = 68 – 72 = – 4

Suppose there are n terms in given AP, we have

\( T_{n}\)  =0

\( \Rightarrow \)   a + (n – 1) d = 0

\( \Rightarrow \)  72 + (n – 1) (-4) = 0

\( \Rightarrow \)  72 – 4n + 4 = 0

\( \Rightarrow \)   4n = 76

\( \Rightarrow \)      n = 19

Hence, the \( 19^{th} \)     term in the given AP is 0.

Q.12: Which term of the \( \frac {5} {6}, 1,  1\frac {1} {6},  1\frac {1} {3},\) ….. is 3 ?

Sol:

In the given AP, first term = \( \frac {5} {6} \) and common difference, \( d = \left ( 1 – \frac {5} {6} = \frac {1} {6} \right ) \)

Let its nth term be 3.

Now, Tn = 3

\( \Rightarrow a + \left ( n – 1 \right )d = 3 \)

\( \Rightarrow \frac {5} {6} + \left ( n – 1 \right ) \times \frac {1} {6} = 3 \)

\( \Rightarrow \frac {2} {3} + \frac {n} {6} = 3 \)

\( \Rightarrow \frac {n} {6} = \frac {7} {3} \)

\( \Rightarrow n = 14 \)

Hence, the 14th term of the given AP is 3.

Q.13: Which term of the AP 21, 18, 15, ……… is -81?

Sol:

The given AP is 21, 18, 15, ……… is -81

First term, a = 21

Common difference, d = \(18 – 21 = -3 \)

Suppose nth term of the given AP is -81.

Then, \(a_{n} = – 81 \)

\(\Rightarrow 21 +  (n – 1) \times (-3) = – 81 \)

\(\Rightarrow -3(n – 1) = -81 – 21 = – 102 \)

\(\Rightarrow n – 1 = 34 \)

\(\Rightarrow n = 35 \)

Hence, the 35th term of the AP is -81.

Q.14: Which term of the AP 3, 8, 13, 18, ………. Will be 55 more than its 20th term?

Sol:

Here a = 3 and d = (8 – 3) = 5

The 20th term is given by

\(T_{20} = a + (20 – 1)d = a + 19d = 3 + 19 \times 5 = 98\)

Therefore, the required term = (98 + 55) = 153

Let this be the nth term.

Then \(T_{n} = 153 \)

\(\Rightarrow\) 3 + (n – 1) = 153

\(\Rightarrow\) 5n = 155

\(\Rightarrow\) n = 31

Hence, the 31st term will be 55 more than its 20th term.

Q.15: Which term of the AP 5, 15, 25, …… will be 130 more than its 31st term?

Sol:

Here a = 5 and d = (15 – 5) = 10

The 31st term is given by

\(T_{31} = a + (31 – 1)d = a + 30d = 5 + 30 \times 10 = 305 \)

Therefore, the required term = (305 + 130) = 453

Let this be the nth term.

Then \(T_{n} = 453 \)

\(\Rightarrow\) 5 + (n – 1) = 453

\(\Rightarrow 5 + (n – 1) \times 10 = 453 \)

\(\Rightarrow\) 10n = 440

\(\Rightarrow\) n = 44

Hence, the 44th term will be 130 more than its 31st term.

Q.16: If the 10th term of an AP is 52 and the 17th term is 20 more than its 13th term, find the AP.

Sol:

In the given AP let the first term = a, and common difference = d

Then, \( T_ {n} \) = a + (n-1) d

\(\Rightarrow\)   \( T_ {10} \) = a + (10 – 1) d,

\( T_ {17} \) = a + (17 – 1) d,

\( T_ {13} \) = a + (13 – 1) d

\( \Rightarrow \)   \( T_ {10} \) = a + 9d,

\( T_ {17} \) = a + 16d,

\( T_ {13} \) = a + 12d

Now, \( T_{10} \)  = 52

\( \Rightarrow \)   a + 9d = 52 – – – (1)

and \( T_{17} \)  = \( T_{13} \)  + 20

\( \Rightarrow \)   a + 16d = a + 12d + 20

\( \Rightarrow \)      4d = 20

\( \Rightarrow \)      d = 5

Putting d = 5 in (1), we get

a + 9 x 5 = 52

\( \Rightarrow \)      a = 52-45

\( \Rightarrow \)      a = 7

Thus, a = 7 and d = 5

So the required AP is 7, 12, 17, 22….

Q.17: Find the middle term of the AP 6, 13, 20,…….., 216.

Sol:

The given AP is 6, 13, 20, ………..216.

First term, a = 6

Common difference, d = 31 – 6 = 7

Suppose there are n terms in the AP. Then,

\(a_{n} = 216\)

\(\Rightarrow 6 + (n – 1) \times 7 = 216\)

\(\Rightarrow 7 (n – 1) = 216 – 6 = 210\)

\(\Rightarrow n – 1 = \frac{210}{7} = 30\)

\(\Rightarrow n = 31 \)

Thus, the given AP contains 31 terms.

Therefore, Middle term of the AP

= \(\left ( \frac{31 + 1}{2} \right ) = 16th \; term\)

Q.18: Find the middle term of the AP 10, 7, 4, …… -62.

Sol:

The given AP is 10, 7, 4, …… -62

First term, a = 10

Common difference, d = 7 – 10 = -3

Suppose there are n terms in the given AP. Then,

\(a_{n} = -62\)

\(\Rightarrow 10 + (n – 1) \times (-3) = -62\)

\(\Rightarrow -3 (n – 1) = -62 – 10 = – 72\)

\(\Rightarrow (n – 1) = -\frac{- 72}{3} = 24\)

\(\Rightarrow n = 25\)

Thus, the given AP contains 25 terms.

Therefore, Middle term of the AP

= \(\left ( \frac{25 + 1}{2} \right )\)

= 13th term

= 10 + (13 – 1) \(\times\) (-3)

= 10 – 36

= -26

Hence, the middle term of the given AP = -26

Q.19: Find the sum of two middle most terms of the AP \(-\frac{4}{3}, -1 , -\frac{2}{3}, ……, 4 \frac {1} {3} \)

Sol:

We know that:

\( T_{1} \) –  (5x + 2) , \( T_{2} \)  – (4x – 1) and \( T_{3} \)  – ( x + 2)

Clearly,

\( T_{2} \)\(T_{1} \)  = \(T_{3} \)\( T_{2} \)

\( \Rightarrow \)  (4x – 1) – (5x + 2) = (x + 2) – (4x – 1)

\( \Rightarrow \) 4x – 1 – 5x – 2 = x + 2 – 4x + 1

\( \Rightarrow \)   -x – 3 = -3x + 3

\( \Rightarrow \)   -x + 3x = 6

\( \Rightarrow \)  2x = 6

\(\Rightarrow\)  x = 3

Hence  x = 3

Q.20: Find the 8th term from the end of the AP 7, 10, 13, ………, 184.

Sol:

Here a = 7, d = (10 – 7) = 3, I = 184

And n = 8

Now, \( n^ {th} \)  term from the end = [I – (n – 1) d]

= [184 – (8 – 1) 3]

= [184 – 7 x 3]

= 184 – 21

= 163

Hence, the \( 8^ {th} \) term from the end is 163.


Practise This Question

 An object is immersed in a fluid. In order that the object becomes invisible, it should