# 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.

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