Practice MCQ Questions and Answer on Progressions

41.

Two A.P.ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂs have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30 terms isth

(A) 11
(B) 3
(C) 8
(D) 5

42.

Which term of the A.P. 92, 88, 84, 80, ...... is 0?

(A) 23
(B) 32
(C) 22
(D) 24

43.

What is the sum of the first 12 terms of an arithmetic progression if the 3^rd term is -13 and the 6^th term is -4?

(A) 67
(B) 45
(C) -30
(D) -48

Solution: $\begin{aligned} T_{3} = a + 2 d = - 13 . . . . . (1) \\ T_{6} = a + 5 d = - 4 . . . . . (2) \\ on solving (1) and (2) \\ d = 3 & a = - 19 \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \end{aligned}$ $S_{12} = \frac{12}{2}$ $[2 (- 19) + (12 - 1) (3)]$ $\begin{aligned} S_{12} = (6) [- 38 + 33] \\ S_{12} = - 30 \end{aligned}$

44.

A bacteria gives birth to two new bacteria in each second and the life span of each bacteria is 5 seconds. The process of the reproduction is continuous until the death of the bacteria. initially there is one newly born bacteria at time t = 0, the find the total number of live bacteria just after 10 seconds :

(A) $\frac{3^{10}}{2}$
(B) 310 - 210
(C) 243 × (35 -1)
(D) 310 - 25

45.

Sum of n terms of the series $\sqrt{2}$ ÃÂÃÂ $+$ ÃÂÃÂ $\sqrt{8}$ ÃÂÃÂ $+$ ÃÂÃÂ $\sqrt{18}$ ÃÂÃÂ $+$ ÃÂÃÂ $\sqrt{32}$ ÃÂÃÂ $+$ ÃÂÃÂ ....... is

(A) $\frac{n (n + 1)}{2}$
(B) $2 n (n + 1)$
(C) $\frac{n (n + 1)}{\sqrt{2}}$
(D) $1$

Solution: The series is given $\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32}$ $+$ ...... $\Rightarrow \sqrt{2} + 2 \sqrt{2} + 3 \sqrt{2}$ $+$ $4 \sqrt{2}$ $+$ ...... Here a = $\sqrt{2}$ and d = $2 \sqrt{2}$ $-$ $\sqrt{2}$ = $\sqrt{2}$ $\begin{aligned} ∴ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ = \frac{n}{2} [2 \sqrt{2} + (n - 1) \sqrt{2}] \\ = \frac{n}{2} [2 \sqrt{2} + \sqrt{2} n - \sqrt{2}] \\ = \frac{n}{2} (\sqrt{2} n + \sqrt{2}) \\ = \frac{n \sqrt{2}}{2} (n + 1) \\ = \frac{n (n + 1)}{\sqrt{2}} \end{aligned}$

46.

The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is

(A) 600
(B) 765
(C) 640
(D) 680

Solution: $\begin{aligned} 1^{s t} Method : \\ 1^{s t} term = 5; \\ 3^{r d} term = 15; \\ Then, d = 5; \\ 16^{t h} term = a + 15 d \\ = 5 + 15 \times 5 = 80 \\ Sum = n \times \frac{(a + l)}{2} \end{aligned}$ $= no . of terms \times \frac{first term + last term}{2}$ $\begin{aligned} = 16 \times \frac{(5 + 80)}{2} \\ = 16 \times \frac{85}{2} \\ = 8 \times 85 \\ = 680 \end{aligned}$ 2nd Method(Thought Process): Sum = number of terms × average of that AP $\begin{aligned} Sum = 16 \times \frac{(5 + 80)}{2} \\ = 16 \times \frac{85}{2} \\ = 8 \times 85 \\ = 680 \end{aligned}$

47.

The 2^nd and 8^th term of an arithmetic progression are 17 and -1 respectively. What is the 14^th term?

(A) -22
(B) -25
(C) -19
(D) -28

Solution: $\begin{aligned} T_{2} = a + d = 17 . . . . . . . (1) \\ T_{8} = a + 7 d = - 1 . . . . . . (2) \\ on solving (1) and (2) \\ d = - 3 & a = 20 \\ T_{14} = a + 13 d \\ = 20 + 13 (- 3) \\ = - 19 \end{aligned}$

48.

If a rubber ball consistently bounces back $\frac{2}{3}$ of the height from which it is dropped, what fraction of its original height will the ball bounce after being dropped and bounced four times without being stopped?

(A) $\frac{16}{81}$
(B) $\frac{16}{27}$
(C) $\frac{4}{9}$
(D) $\frac{37}{81}$

Solution: Each time the ball is dropped and it bounces back, it reaches $\frac{2}{3}$ of the height it was dropped from. After the first bounce, the ball will reach $\frac{2}{3}$ of the height from which it was dropped - let us call it the original height. After the second bounce, the ball will reach $\frac{2}{3}$ of the height it would have reached after the first bounce. So, at the end of the second bounce, the ball would have reached $\frac{2}{3}$ × $\frac{2}{3}$ of the original height = $\frac{4}{9}$ th of the original height. After the third bounce, the ball will reach $\frac{2}{3}$ of the height it would have reached after the second bounce. So, at the end of the third bounce, the ball would have reached $\frac{2}{3}$ × $\frac{2}{3}$ × $\frac{2}{3}$ = $\frac{8}{27}$ th of the original height. After the fourth and last bounce, the ball will reach $\frac{2}{3}$ of the height it would have reached after the third bounce. So, at the end of the last bounce, the ball would have reached $\frac{2}{3}$ × $\frac{2}{3}$ × $\frac{2}{3}$ × $\frac{2}{3}$ of the original height = $\frac{16}{81}$ of the original height.

49.

The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by $\frac{l^{2} - a^{2}}{k - (l + a)}$ ÃÂÃÂ then k = ?

(A) S
(B) 2S
(C) 3S
(D) None of these

Solution: $\begin{aligned} S = \frac{n}{2} (l + a) \\ l = a + (n - 1) d \\ d = \frac{l^{2} - a^{2}}{k - (l + a)} and also d = \frac{l - a}{n - 1} \\ ∴ \frac{l - a}{n - 1} = \frac{(l + a) (l - a)}{k - (l + a)} \\ \Rightarrow \frac{1}{n - 1} = \frac{l + a}{k - (l + a)} \\ \Rightarrow k - (l + a) = (n - 1) (l + a) \\ \Rightarrow k = (n - 1) (l + a) + (l + a) \\ \Rightarrow k = (l + a) (n - 1 + 1) \\ \Rightarrow k = n (l + a) \\ \Rightarrow k = 2 \times \frac{n}{2} (l + a) {∴ \frac{n}{2} (l + a) = S} \\ \Rightarrow k = 2 \times S \\ \Rightarrow k = 2 S \end{aligned}$

50.

What is the sum of the first 13 terms of an arithmetic progression if the first term is -10 and last term is 26?

(A) 104
(B) 140
(C) 84
(D) 98

Solution: $\begin{aligned} S_{n} = \frac{n}{2} [a + l] \\ S_{13} = \frac{13}{2} [- 10 + 26] \\ S_{13} = \frac{13}{2} [16] \\ S_{13} = 104 \end{aligned}$

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Practice MCQ Questions and Answer on Progressions

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