# RS Aggarwal Class 10 Solutions Chapter 11 - Arithmetic Progressions Ex 11F (11.6)

## RS Aggarwal Class 10 Chapter 11 - Arithmetic Progressions Ex 11F (11.6) Solutions Free PDF

Q.1: The first and the last term of an AP are 5 and 45 respectively. If the sum of all its terms is 400, then find the common difference and the number of terms.

Sol:

Suppose there are n terms in the AP.

Here, a = 5, l = 45 and $S_{n} = 400$

$S_{n} = 400 = \frac{n}{2} (5 + 45)$

$\Rightarrow \frac{n}{2} (50) = 400$

$\Rightarrow n = 16$

Thus, there are 16 terms in the AP.

Let d be the common difference of the AP

$Therefore, a_{16} = 45$

$\Rightarrow 5 + (16 – 1) \times d = 45$

$\Rightarrow 15 d = 40$

$\Rightarrow d = \frac{8}{3}$

Hence the common difference of the AP is $\frac{8}{3}$.

Q.2: In an AP, the first term is 22, nth term is -11 and sum of first n terms is 66. Find n and hence find the common difference.

Sol:

Here a = 22, $T_{n} = -11 \;\; and \;\; S_{n} = 66$

Let d be the common difference of the given AP.

Then $T_{n} = -11$

$\Rightarrow a + (n – 1)d = 22 + (n – 1)d = -11$

$\Rightarrow (n – 1)d = – 33$       …………(i)

The sum of n terms of an AP is given by

$S_{n} = \frac{n}{2} (2a + (n – 1) d) = 66$

$\Rightarrow \frac{n}{2} (2 \times 22 + (-33)) = \left ( \frac{n}{2} \times 11\right ) = 66$

$\Rightarrow n = 12$

Putting the value of n in (i), we get:

$11d = -33$

$\Rightarrow d = -3$

Thus, n = 12 and d = -3

Q.3: The 12th term of an AP is -13 and the sum of its first four terms is 24. Find the sum of its first 10 terms.

Sol:

Let a be the first term and d be the common difference of the AP.

Then,

$a_{12} = – 13$

$\Rightarrow a + 11 d = – 13$             …………(i)

Also,

$S_{4} = 24$

$\Rightarrow \frac{4}{2} (2a + 3d) = 24$

$\Rightarrow (2a + 3d) = 12$            ………….. (ii)

Solving (i) and (ii) we get:

$2(- 13 – 11d) + 3d = 12$

$\Rightarrow – 26 – 22d + 3d = 12$

$\Rightarrow – 19 d = 12 + 26 = 38$

$\Rightarrow d = -2$

Putting d = -2 in (i) we get:

$\Rightarrow a + 11 ( – 2) = – 13$

$\Rightarrow a = -13 + 22 = 9$

Therefore, Sum of its first 10 terms, $S_{10}$

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

$= 5 \times (18 – 18)$

$= 5 \times 0$

= 0

Hence, the required sum is 0.

Q.4: The sum of the first 7 terms of an AP is 182. If its 4th and 17th terms are in the ratio 1 : 5, find the AP.

Sol:

Let a be the first term and d be the common difference of the AP.

$Therefore, S_{7} = 182$

$\Rightarrow \frac{7}{2} (2a + 6d) = 182$

$\Rightarrow a + 3d = 26$                …………..(i)

Also,

$a_{4} : a_{17} = 1 : 5$                   (given)

$\Rightarrow \frac{a + 3d}{a + 16d} = \frac{1}{5}$

$\Rightarrow 5a + 15 d = a + 16d$

$\Rightarrow d = 4a$                    ……………..(ii)

Solving (i) and (ii), we get

$a + 3 \times 4 a = 26$

$\Rightarrow 13 a = 26$

$\Rightarrow a = 2$

Putting a = 2 in (ii), we get

$d = 4 \times 2 = 8$

Hence, the required AP is 2, 10 , 18, 26, ……….

Q.5: The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first term and the common difference of the AP.

Sol:

Here a= 4 , d = 7 and l = 81

Let the nth term be 81

Then $T_{n} = 81$

$\Rightarrow a + (n – 1)d = 4 + (n – 1)7 = 81$

$\Rightarrow (n – 1)7 = 77$

$\Rightarrow (n – 1) = 11$

$\Rightarrow n = 12$

Thus, there are 12 terms in the AP.

The sum of n terms of an AP is given by

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

$S_{12} = \frac{12}{2} (4 + 81) = 6 \times 85 = 510$

Thus, the required sum is 510.

Q.6: The sum of the first 7 terms of an AP is 49 and the sum of its first 17 terms is 289. Find the sum of its first n terms.

Sol:

Let a be the first term and d be the common difference of the given AP.

Then we have:

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

$S_{7} = \frac{7}{2} \left [ 2a + 6d \right ] = 7 \left [ a + 3d \right ]$

$S_{17} = \frac{17}{2} \left [ 2a + 6d \right ] = 17 \left [ a + 3d \right ]$

However, $S_{7} = 49 \;\; and \;\;S_{17} = 289$

However, $S_{7} = 49 \;\; and \;\;S_{17} = 289$

Now, $7 \left [ a + 3d \right ] = 49$

$\Rightarrow a + 3d = 7$            ………..(i)

Also, $17 \left [ a + 8d \right ] = 289$

$\Rightarrow a + 8 d = 17$            ………..(ii)

Subtracting (i) from (ii), we get:

5d = 10

$\Rightarrow d = 2$

Putting d = 2 in (i), we get:

a + 6 = 7

$\Rightarrow a = 1$

Thus, a = 1 and d = 2

Therefore, Sum of n terms of AP = $\frac{n}{2} \left [ 2 \times 1 + (n -1) \times 2 \right ] = n \left [ 1 + (n – 1) \right ] = n^{2}$

Q.7: Two APs have the same common difference. If the first terms of these APs are 3 and 8 respectively, find the difference between the sums of their first 50 terms.

Sol:

Let $a_{1} \;\; and \;\; a_{2}$ be the first terms of the two AP.

Here, $a_{1} = 8$ and $a_{2} = 3$

Suppose d be the common difference of two AP.

Let $S_{50} \;\; and \;\; {S}’_{50}$ denotes the sum of their first 50 terms.

$Therefore, S_{50} – {S}’_{50} = \frac{50}{2} \left [ 2 a_{1} + (50 – 1) d\right ] – \frac{50}{2} \left [ 2 a_{2} + (50 – 1) d\right ]$

$= 25 \left [ 2 \times 8 + 49 d\right ] – 25 \left [ 2 \times 3 + 49 d\right ]$

$= 25 \times (16 – 6)$

= 250

Hence, the required difference between the two sums is 250.

Q.8: The sum of first 10 terms of an AP is -150 and the sum of its next 10 terms is -550. Find the AP.

Sol:

Let a be the first term and d be the common difference of the AP.

Then,

$S_{10} = -150$                              (Given)

$\Rightarrow \frac{10}{2} (2a + 9d) = -150$

$\Rightarrow 5 (2a + 9d) = -150$

$\Rightarrow (2a + 9d) = -30$                …………….(i)

It is given that the sum of its next 10 terms is – 550

Now,

$S_{20}$ = Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms = – 150 + (- 550 ) = -700

$Therefore, S_{20} = -700$

$\Rightarrow \frac{20}{2} (2a + 19d) = – 700$

$\Rightarrow 10 (2a + 19d) = – 700$

$\Rightarrow 2a + 19d = – 70$             ……………….(ii)

Subtracting (i) from (ii), we get

$2a + 19d – (2a + 9d) = – 70 – (-30)$

$\Rightarrow 10 d = -40$

$\Rightarrow d = -4$

Putting d = -4 in (i), we get

$2a + 9 \times (- 4) = – 30$

$\Rightarrow 2a = – 30 + 36 = 6$

$\Rightarrow a = 3$

Hence, the required AP is 3, -1, -5, -9, ……

Q.9: The 13th term of an AP is 5 times its 3rd term. If its 5th term is 16, find the sum of its first 10 terms.

Sol:

Let a be the first term and d be the common difference of the AP. Then,

$a_{13} = 4 \times a_{3}$                       (Given)

$\Rightarrow a + 12 d = 4 (a + 2d)$

$\Rightarrow a + 12 d = 4 a + 8d$

$\Rightarrow 3a = 4d$           …………..(i)

Also,

$a_{5} = 16$                   (Given )

$\Rightarrow a + 4d = 16$                   …………(ii)

Solving (i) and (ii), we get

$a + 3a = 12$

$4a = 12$

$a = 3$

Putting a = 4 in (i) we get:

$4d = 3 \times 4 = 12$

$\Rightarrow d = 3$

Using the formula,

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

$S_{10} = \frac{10}{2} \left [ 2 \times 4 + (10 – 1) \times 3 \right ]$

$= 5 \left [ 8 + 27 \right ]$

$= 5 \left [35 \right ]$

$= 175$

Hence the required sum is 175.

Q.10: The 16th term of an AP is 5 times its 3rd term. If its 10th term is 41, find the sum of its first 10 terms.

Sol:

Let a be the first term and d be the common difference of the AP.

Then,

$a_{16} = 5 \times a_{3}$                       (Given)

$\Rightarrow a + 15 d = 5 (a + 2d)$

$\Rightarrow a + 12 d = 5a + 10d$

$\Rightarrow 4a = 5d$           …………..(i)

Also,

$a_{10} = 41$                   (Given )

$\Rightarrow a + 9d = 41$                   …………(ii)

Solving (i) and (ii), we get

$a + 9 \times \frac{4}{5}a = 41$

$\Rightarrow \frac{5a + 36a}{5} = 41$

$\Rightarrow \frac{41a}{5} = 41$

$a = 5$

Putting a = 5 in (i) we get:

$5d = 4 \times 5 = 20$

$\Rightarrow d = 4$

Using the formula,

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

$S_{15} = \frac{15}{2} \left [ 2 \times 5 + (15 – 1) \times 4 \right ]$

$= \frac{15}{2} \left [ 10 + 56 \right ]$

$= \frac{15}{2} \left [66 \right ]$

$= 495$

Hence the required sum is 495.

Q.11: An AP 5, 12, 19, … has 50 terms. Find its last term. Hence, find the sum of its last 15 terms.

Sol:

The given AP is 5, 12, 19, ………..

Here, a = 5, d = 12 – 5 = 7 and n = 50.

Since there are 50 terms in the AP, so the last term of the AP is $a_{50}$.

l = $a_{50}$ = 5 + (50 – 1) $\times$ 7              [ $a_{n} = a + (n – 1)d$ ]

= 5 + 343

= 348

Thus, the last term of the AP is 348.

Now,

Sum of the last 15 term of the AP

= $S_{50} – S_{35}$

$= \frac{50}{2} \left [ 2 \times 5 + (50 – 1) \times 7 \right ] – \frac{35}{2} \left [ 2 \times 5 + (35 – 1) \times 7 \right ]$

= $= \frac{50}{2} \left [ 10 + 343 \right ] – \frac{35}{2} \left [ 10 + 238 \right ]$

$= (25 \times 353) – (35 \times 124)$

$= 4485$

Hence, the required sum is 4485.

Q.12: An AP 8, 10, 12, … has 60 terms. Find its last term. Hence, find the sum of its last 10 terms.

Sol:

The given AP is 8, 10, 12, …………

Here, a = 8 , d = 10 – 8 = 2 and n = 60

Since there are 60 terms in the AP, so the last term of the AP is $a_{50}$.

l = $a_{50}$ = 8 + (60 – 1) $\times$ 2     [ $a_{n} = a + (n – 1)d$ ]

Thus, the last term of the AP is 126.

Now,

Sum of the last 10 terms of the AP

= $S_{60} – S_{50}$

$= \frac{50}{2} \left [ 2 \times 5 + (50 – 1) \times 7 \right ] – \frac{35}{2} \left [ 2 \times 5 + (35 – 1) \times 7 \right ]$

= $= \frac{50}{2} \left [ 10 + 343 \right ] – \frac{35}{2} \left [ 10 + 238 \right ]$

$= (25 \times 353) – (35 \times 124)$

$= 4485$

Hence, the required sum is 4485.

Q.13: The sum of the 4th and 8th terms of AP is 24 and the sum of its 6th and 10th terms is 44. Find the sum of its first 10 terms.

Sol:

Let a be the first term and d be the common difference of the AP.

$Therefore, a_{4} + a_{8} = 24$            ( given)

$\Rightarrow$ (a + 3d) + (a + 7d) = 24

$\Rightarrow$ 2a + 10d = 24

$\Rightarrow$ a + 5d = 12         …………(i)

Also,

$Therefore, a_{6} + a_{10} = 44$            ( given)

$\Rightarrow$ (a + 5d) + (a + 9d) = 44

$\Rightarrow$ 2a + 14d = 44

$\Rightarrow$ a + 7d = 22         …………(i)

Subtracting (i) from (ii), we get

(a + 7d) – (a + 5d) = 22 -12

$\Rightarrow 2d = 10$

$\Rightarrow d = 5$

Putting d = 5 in (i), we get

a + 5$\times$ 5 = 12

$\Rightarrow$ a = 12 – 25 = -13

Using the formula, $S_{n} = \frac{n}{2} \left [ 2a + (n – 1)d \right ]$, we get

$= \frac{10}{2} \left [ 2 \times (-13) + (10 – 1) \times 5 \right ]$

= $= 5 \times (- 26 + 45 )$

$= 5 \times 19$

$= 95$

Hence, the required sum is 95.

Q.14: The sum of first m terms of an AP is ( $4m^{ 2 }$ – m ). If its nth term is 107, find the value of n. Also, find the 21st term of this AP.

Sol:

Let $S_{m}$ denotes the sum of first m terms of the AP.

Then,

$S_{m} = 4m^{2} – m$

$\Rightarrow S_{m-1} = 4(m – 1)^{2} – (m – 1)$

$= 4(m^{2} – 2m + 1) – (m – 1)$

$= 4m^{2} – 9m + 5$

Suppose $a_{m}$ denotes the mth term of the AP.

$Therefore, a_{m}$ =  $S_{m}$$S_{m – 1}$

$= (4m^{2} – m) – (4m^{2} – 9m + 5 )$

= 8m – 5                      ……………….(i)

Now,

$a_{n}$ = 107                     (given)

$\Rightarrow 8n – 5 = 107$                     [ from (i) ]

$\Rightarrow 8n = 107 + 5 = 112$

$\Rightarrow n = 14$

Thus, the value of n is 14.

Putting m = 21 in (i), we get

$a_{21} = 8 \times 21 – 5 = 168 – 5 = 163$

Hence, the 21st term of the AP is 163.

#### Practise This Question

What made the kulhs water irrigation system used in Himachal Pradesh better than the trivial irrigation system.