# RS Aggarwal Solutions Class 10 Ex 11A

Q.1: Find the 6th term from the end of the AP 17, 14, 11, …….., (-40) .

Sol:

Here a = 17, d = (14 – 17) = – 3, I = -40

And n = 6

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

= [- 40 – (6 – 1) (-3)]

= [- 40 + 5 x 3]

= – 40+15

= – 25

Hence, the 6th term from the end is   – 25.

Q2: Is 184 a term of the AP 3, 7, 11, 15, ……. ?

Sol:

The given AP is 3, 7, 11, 15, …….

Here, a = 3 and d = 7 – 3 = 4

Let nth term of the given AP be 184.

Then,

$a_{n} = 184$

$\Rightarrow 3 + (n – 1) \times 4 = 184$

$\Rightarrow 4n – 1 = 184$

$\Rightarrow 4n = 185$

$\Rightarrow n = \frac{184}{4} = 46 \frac{1}{4}$

But the number of terms cannot be a fraction.

Hence, 184 is not a term of the given AP.

Q.3: Is – 150 a term of the AP 11, 8, 5, 2, …… ?

Sol:

The given AP is 11, 8, 5, 2, ……

Here, a = 11 and d = 8 – 11 = – 3

Let nth term of the given AP be 150.

Then,

$a_{n} = 150$

$\Rightarrow 11 + (n – 1) \times (-3) = 150$

$\Rightarrow – 3n + 14 = – 150$

$\Rightarrow – 3n = – 164$

$\Rightarrow n = \frac{164}{3} = 54 \frac{2}{3}$

But the number of terms cannot be a fraction.

Hence, -150 is not a term of the given AP.

Q.4: Which term of the AP 121, 117, 113, ….. is its first negative term?

Sol:

The given AP is 121, 117, 113, …..

Here, a = 121 and d= 117 – 121 = -4

Let the nth term of the given AP be the first negative term. Then,

$a_{n} < 0$

$\Rightarrow 121 + (n – 1) \times (-4) < 0$

$\Rightarrow 125 – 4n < 0$

$\Rightarrow -4n < -125$

$\Rightarrow n > \frac{125}{4} = 31\frac{1}{4}$

$Therefore, n = 32$

Hence, the 32nd term is the first negative term of the given AP.

Q.5: Which term of the AP is $20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, …….$ is its negative term?

Sol:

The given AP is $20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, …….$

Here, a = 20 and d = $19\frac{1}{4} – 20 = \frac{77}{4} – 20 = -\frac{3}{4}$

Let the nth term of the given AP be the first negative term. Then,

$a_{n} < 0$

$\Rightarrow 20 + (n – 1) \times \left ( -\frac{3}{4} \right ) < 0$

$\Rightarrow 20 + \frac{3}{4} – \frac{3}{4}n < 0$

$\Rightarrow \frac{83}{4} – \frac{3}{4}n < 0$

$\Rightarrow -\frac{3}{4}n < – \frac{83}{4}$

$\Rightarrow n > \frac{83}{3} = 27 \frac{2}{3}$

$Therefore, n = 28$

Hence, the 28th term is the first negative term of the given AP.

Q.6: The 7th term of an AP is -4 and its 13th term is -16. Find the AP.

Sol:

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

Then, $T_ {n}$ = a + (n – 1) d

$\Rightarrow$   $T_ {7}$ = a + (7 – 1)d, and $T_{13}$  = a + (13 – 1)d

$\Rightarrow$   $T_ {7}$ = a + 6d, $T_ {13}$ = a + 12d

Now, $T_ {7}$ = -4

$\Rightarrow$ a + 6d = -4 – – – (1)

$T_{13}$ = -16

$\Rightarrow$   a + 12d = -16 – – – (2)

Subtracting (1) from (2), we get

$\Rightarrow$ 6d = -12 d = -2

Putting d = -2 in (1), we get

a + 6 (-2) = -4

$\Rightarrow$ a – 12 = -4

$\Rightarrow$   a = 8

Thus, a = 8, and d = -2

So the required AP is 8, 6, 4, 2, 0   …

Q.7: The 4th term of an AP is 0. Prove that its 25th term is triple its 11th term.

Sol:

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

Then, $T_ {n}$ = a + (n – 1) d

Now, $T_ {4}$ = a + (4 – 1) d

$\Rightarrow$ a + 3 d = 0……..(i)

$\Rightarrow$ a = -3d

Again, $T_ {11}$ = a + (11 – 1)d =  a + 10d

= -3d + 10 d = 7d             [Using (i) ]

Also, $T_ {25}$ = a + (25 – 1)d =  a + 24d = -3d + 24d = 21d                       [Using (i) ]

i.e. $T_{25} = 3 \times 7d = 3 \times T_{11}$

Hence, 25th term is triple its 11th term.

Q.8: The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.

Sol:

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

$a_{8} = 0$

$\Rightarrow a + (8 – 1)d = 0$

$\Rightarrow a + 7 d = 0$

$\Rightarrow a = -7d$ …………..(i)

Now,

$\Rightarrow \frac{a_{38}}{a_{18}} = \frac{a + (38 – 1)d}{a + (18 – 1)d}$

$\Rightarrow \frac{a_{38}}{a_{18}} = \frac{ – 7d + 37 d} {- 7d + 17d}$

$\Rightarrow \frac{a_{38}}{a_{18}} = \frac{30d } {10 d} = 3$

$a_{38} = 3 \times a_{18}$

Hence, the 38th term of the AP is triple its 18th term.

Q.9: The 4th term of an AP is 11. The sum of the 5th and 7th term of this AP is 34. Find its common difference.

Sol:

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

$a_{4} = 11$

$\Rightarrow a + (4 – 1)d = 11$

$\Rightarrow a + 3 d = 11$        …………(i)

Now,

$a_{5} + a_{7} = 34$             (Given)

$\Rightarrow (a + 4d) + (a + 6d) = 34$

$\Rightarrow 2a + 10d = 34$

$\Rightarrow a + 5d = 17$            ……….(ii)

From (i) and (ii) we get

11 – 3d + 5d = 17

$\Rightarrow 2d = 17 – 11 = 6$

$\Rightarrow d = 3$

Hence, the common difference of the AP is 3.

Q.10: The 9th term of an AP is -32 and the sum of its 11th and 13th term is -94. Find the common difference of the AP.

Sol:

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

$a_{4} = 11$

$\Rightarrow a + (9 – 1)d = – 32$

$\Rightarrow a + 8 d = -32$        …………(i)

Now,

$a_{11} + a_{13} = -94$             (Given)

$\Rightarrow (a + 10d) + (a + 12d) = -94$

$\Rightarrow 2a + 22d = -94$

$\Rightarrow a + 11d = -47$            ……….(ii)

From (i) and (ii) we get

– 32 – 8d + 11d = -47

$\Rightarrow 3d = -47 + 32 = -15$

$\Rightarrow d = -5$

Hence, the common difference of the AP is -5.

Q.11: Determine the nth term of the AP whose 7th term is -1 and 16th term is 17.

Sol:

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

$a_{ 7 }$ = -1

→ a + (7 – 1) d = -1        $\left [ a_{ n } = a + (n – 1)d \right ]$

→ a + 6d = -1       … (1)

Also,

$a_{ 16 }$ = 17

→ a + 15d = 17              …(2)

From (1) and (2), we get

-1 – 6d + 15d = 17

→ 9d = 17 + 1 = 18

→ d = 2

Putting d = 2 in (1), we get

a + 6 x 2 = -1

→ a = -1 – 12 = -13

Therefore, $a_{ n } = a + (n – 1)d$

= -13 + (n – 1) x 2

= 2n – 15

Hence, the nth term of the AP is (2n – 15)

Q.12: If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term.

Sol:

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

4 x $a_{ 4 } = 18 \times a_{ 18 }$     (Given)

→ 4 (a + 3d) = 18 (a + 17d)     $\left [ a_{ n } = a + (n – 1)d \right ]$

→ 2 (a + 3d) = 9 (a + 17d)

→ 2a + 6d = 9a + 153d

→ 7a = -147d

→ a = -21d

→ a + 21d = 0

→ a + (22 – 1)d = 0

$a_{ 22 }$ = 0

Hence, the 22nd term of the AP is 0.

Q.13: If 10 times the 10th term of an AP is equal to 15 times the 15th term, show that its 25th term is zero.

Sol:

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

10 x $a_{ 10 } = 15 \times a_{ 15 }$ (Given)

→ 10 (a + 9d) = 15 (a + 14d)   $\left [ a_{ n } = a + (n – 1)d \right ]$

→ 2 (a + 9d) = 3(a + 14d)

→ 2a + 18d = 3a + 42d

→ a = -24d

→ a + 24d = 0

→ a + (25 – 1) d = 0

$a_{ 25 }$ = 0

Hence, the 25th term of the AP is 0.

Q.14: Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.

Sol:

Let the common difference of the AP be d.

First term, a = 5

Now,

$a_{ 1 } + a_{ 2 } + a_{ 3 } + a_{ 4 } = \frac{ 1 }{ 2 } (a_{ 5 } + a_{ 6 } + a_{ 7 } + a_{ 8 })$            (Given)

→ a + (a + d) + (a + 2d) + (a + 3d) = $\frac{ 1 }{ 2 }$ [ (a + 4d) + ( a + 5d) + (a + 6d) + (a + 7d)]

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

→ 4a + 6d = $\frac{ 1 }{ 2 }$ (4a + 22d)

→ 8a + 12d = 4a + 22d

→ 22d – 12d = 8a – 4a

→ 10d = 4a

→ d = $\frac{ 2 }{ 5 }a$

→ d = $\frac{ 2 }{ 5 } \times 5 = 2$   (a = 5)

Hence, the common difference of the AP is 2.

Q.15: The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP.

Sol:

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

$a_{ 2 } + a_{ 7 } = 30$

Therefore, (a + d) + (a + 6d) = 30     $\left [ a_{ n } = a + (n – 1)d \right ]$

→ 2a + 7d = 30              ….(1)

Also,

$a_{ 15 } = 2a_{ 8 } – 1$          (Given)

→ a + 14d = 2 (a + 7d) – 1

→ a + 14d = 2a + 14d – 1

→ -a = -1

→ a = 1

Putting a = 1 in (1), we get

2 x 1 + 7d = 30

→ 7d = 30 – 2 = 28

→ d = 4

So,

$a_{ 2 }$ = a + d = 1 + 4 = 5

$a_{ 3 }$ = a +2d = 1 + 2 x 4 = 9,…

Hence, the AP is 1, 5, 9, 13,…

Q.16: For what value of n, the nth terms of the arithmetic progressions 63, 65, 67,… and 3, 10, 17,.. are equal?

Sol:

Let the nth term of the given progressions be $t_{ n }$ and $T_{ n }$ ,respectively.

The first AP is 63, 65, 67,…

Let its first term be a and the common difference be d.

Then a = 63, and d = (65 – 63) = 2

So, its $n^{th}$ term is given by

$T_{ n }$ = A + (n -1) D

→ 3 + (n -1) x 7

→ 7n – 4

Now, $t_{ n }$ = $T_{ n }$

→ 61 + 2n = 7n – 4

→ 65 = 5n

→ n = 13

Hence, the 13th terms of the AP’s are the same.

Q.17: The 17th term of an AP is 5 times more than twice its 8th term. If the 11th term of the AP is 43, find its nth term.

Sol:

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

$a_{ 17 } = 2a_{ 8 } + 5$                   (Given)

Therefore, a + 16d = 2 (a + 7d) + 5   $\left [ a_{ n } = a + (n – 1)d \right ]$

→ a + 16d = 2a + 14d + 5

→ a – 2d = -5        …(1)

Also,

$a_{ 11 }$ = 43               (Given)

→ a + 10d = 43              …(2)

From (1) and (2), we get

-5 + 2d + 10d = 43

→ 12d = 43 + 5 = 48

→ d = 4

Putting d = 4 in (1), we get

a – 2 x 4 = -5

→ d = 4

Therefore, $a_{ n } = a + (n – 1)d$

= 3 + (n – 1) x 4

= 4n – 1

Hence, the nth term of the AP is (4n – 1)

#### Practise This Question

The following reaction is an example of a
2Pb(NO3)2(s)Heat−−2PbO(s)+4NO2(g)+O2(g)
I. Combination reaction
II.
Decomposition reaction
III. Endothermic reaction
IV. Exothermic reaction