 RS Aggarwal Class 10 Solutions Chapter 11 - Arithmetic Progressions

RS Aggarwal Class 10 Chapter 11 - Arithmetic Progressions Solutions Free PDF

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.

Access CBSE Sample Paper for class 10 Maths here.

Access NCERT Book for class 10 Maths here.

Practise This Question

Identify the following contraceptives in order from left to right. 