Sequences And Series(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

SEQUENCES AND SERIES MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21.

If $(a - b)^{2}, (a^{2} + b^{2})$ are the first two terms of an AP, then which of the following will be the next term?

2ab
$(a + b)^{2}$
- 2ab
$- (a + b)^{2}$

Discuss Question

Answer: Option B. ->

(a + b)^{2}

:
B
Given,

a_{1} = (a - b)^{2}

a_{1} = a^{2} + b^{2} - 2 a b

and

a_{2} = a^{2} + b^{2}

Now,

d = a_{2} - a_{1}

d = a^{2} + b^{2} - a^{2} - b^{2} + 2 a b

\Rightarrow d = 2 a b

The next term

a_{3} = a_{2} + d

a_{3} = a^{2} + b^{2} + 2 a b = (a + b)^{2}

∴ N e x t t e r m (a_{3}) = (a + b)^{2}

Question 22.

If $m \neq n$ and the sequences $p, a, b, q$ and $p, m, n, q$ each are in AP, then $\frac{b - a}{n - m}$ is ___.

$\frac{2}{3}$
$\frac{3}{2}$
1
$\frac{3}{4}$

Discuss Question

Answer: Option C. -> 1
:
C
Given,

p, a, b, q

are in AP.

∴ 2 a = p + b

\Rightarrow 2 a - b = p . . . . (i)

Also, 2 b = a + q .

\Rightarrow 2 b - a = q . . . . (i i)

Also given that

p, m, n, q

are in AP.

∴ 2 m = p + n

\Rightarrow 2 m - n = p . . . (i i i)

Also, 2 n = m + q .

\Rightarrow 2 n - m = q . . . (i v)

From eqns.

(i)

(i i i),

we get

2 a - b = 2 m - n . . . . (v)

From eqns.

(i i)

(i v),

we get

2 b - a = 2 n - m . . . . (v i)

Subtracting

(v i)

from

(v),

we get

2 a - b - (2 b - a) = 2 m - n - (2 n - m) .

\Rightarrow 2 a - b - 2 b + a = 2 m - n - 2 n + m

3 a - 3 b = 3 m - 3 n

b - a = n - m

\Rightarrow \frac{b - a}{n - m} = 1

Question 23.

If the $5^{t h}$ term of a G.P. containing positive terms is $\frac{1}{3}$ and $9^{t h}$ term is $\frac{16}{243}$ , then the $4^{t h}$ term will be

$\frac{3}{4}$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{2}{5}$

Discuss Question

Answer: Option B. ->

\frac{1}{2}

:
B

$T_{5}$ = $a r^{4}$ = $\frac{1}{3}$ .......(i)

and $T_{9}$ = $a r^{8}$ = $\frac{16}{243}$ .......(ii)

Solving (i) and (ii), we get r = $\frac{2}{3}$ and a = $\frac{27}{16}$

Now $4^{t h}$ term = $a r^{3}$ = $\frac{3^{3}}{2^{4}} . \frac{2^{3}}{3^{3}}$ = $\frac{1}{2}$

Question 24.

The number of terms between 1 to 1000 divisible by 7 are ___.

Discuss Question

Answer: Option A. -> 142
:
A
The required sequence will be 7,14, 21, 28, . . . . 994.
Now, the number of terms divisible by 7 will be
994 = 7 + (n - 1)7
[

∵ a_{n}

= a + (n - 1) d]
Where,
a = 7,
d = 14 -7 = 7 and
n = Number of terms.

\Rightarrow 994 = 7 + 7 n - 7

\Rightarrow 994 = 7 n

\Rightarrow n = \frac{994}{7}

\Rightarrow n = 142

Hence, the number of terms between 1 to 1000 that are divisible by 7 are 142.

Question 25.

If the $t^{th}$ term of an AP is $s$ and $s^{th}$ term of the same AP is $t,$ then $a_{n}$ is ___.

$t + s + n$
$t + s - n$
$t - s + n$
$t - s - n$

Discuss Question

Answer: Option B. ->

t + s - n

:
B
Given,

a_{t} = a + (t - 1) d = s, . . . (i)

a_{s} = a + (s - 1) d = t . . . (i i)

\Rightarrow (t - s) = a + (s - 1) d - a - (t - 1) d

\Rightarrow t - s = (s - t) d [from eqns. (i) and (ii)]

\Rightarrow (t - s) = - (t - s) d

\Rightarrow d = - 1

From

(i),

we get

a + (t - 1) (- 1) = s .

\Rightarrow a = s + t - 1

a_{n} = a + (n - 1) d

\Rightarrow a_{n} = s + t - 1 + (n - 1) \times (- 1) = s + t - 1 - n + 1 = s + t - n

Question 26.

The number which should be added to the number added to the numbers 2, 14, 62 so that the resulting numbers may be in G.P. is

Discuss Question

Answer: Option B. -> 2
:
B

Suppose that the added number be x

then x + 2, x + 14, x + 62 be in G.P.

Therefore $(x + 14)^{2}$ = (x + 2)(x + 62)

$\Rightarrow x^{2} + 196 + 28 x$ = $x^{2} + 64 x + 124$

$\Rightarrow 36 x$ = 72 $\Rightarrow x$ = 2

Trick : (a) Let 1 is added, then the numbers will be 3, 15, 63 which are obviously not in G.P.

(b) Let 2 is added, then the numbers will be 4, 16, 64 which are obviously in G.P.

Question 27.

If 8 times the $8^{th}$ term of an AP is equal to 15 times the $15^{th}$ term of the AP, then the $23^{rd}$ term of the AP is ___.

Discuss Question

Answer: Option C. -> 0
:
C
Given,

8 a_{8} = 15 a_{15} .

Since the

n^{th}

term of an AP of first term

a

and common difference

d

is given by

a_{n} = a + (n - 1) d,

we have

8 [a + (8 - 1) d] = 15 [a + (15 - 1) d] . \Rightarrow 8 (a + 7 d) = 15 (a + 14 d) \Rightarrow 8 a + 56 d = 15 a + 210 d \Rightarrow 7 a + 154 d = 0 \Rightarrow a + 22 d = 0 \Rightarrow a + (23 - 1) d = 0

Hence, the

23^{rd}

term of the AP is zero.

Question 28.

In a flag race, a pole is placed at the starting point, which is 10m from the first flag and the other flags are placed 6m apart in a straight line. There are 10 flags in the line. Each competitor starts from the pole, picks up the nearest flag and comes back to the pole and runs for the next flag and continues the same way until all the flags are on the pole. The total distance covered is ____.

740 m
760 m
820 m
860 m

Discuss Question

Answer: Option A. -> 740 m
:
A
We have,

d_{1}

= distance travelled to pick up the first flag
= 10 + 10 = 2

\times

10 = 20 m

d_{2}

= distance travelled to pick up the second flag
= 2

\times

(10 + 6) = 32m

d_{3}

= distance travelled to pick up the third flag
= 2

\times

(10 + 6 + 6)
= 44 m

d_{10}

= distance travelled to pick up 10th flag
= 2

\times

[10 + (10 - 1)

\times

6]m

∴

Total distance travelled

= d_{1} + d_{2} + d_{3} + \dots + d_{10} = \frac{10}{2} [2 \times 20 + (10 - 1) 12] = 5 (40 + 108) = 5 \times 148

= 740m

Question 29.

If a, b, c are $p^{t h}$ , $q^{t h}$ , and $r^{t h}$ terms of a G.P., then $(\frac{c}{b})^{p} (\frac{b}{a})^{r} (\frac{a}{c})^{q}$ is equal to

1
$a^{p} b^{q} c^{r}$
$a^{q} b^{r} c^{p}$
$a^{r} b^{p} c^{q}$

Discuss Question

Answer: Option A. -> 1
:
A

a = $A R^{p - 1}$ , b = $A R^{q - 1}$ , c = $A R^{r - 1}$

$∴ (\frac{c}{b})^{p} (\frac{b}{a})^{r} (\frac{a}{c})^{q}$ = $(\frac{A R^{p - 1}}{A R^{q - 1}})^{p} (\frac{A R^{q - 1}}{A R^{p - 1}})^{r} (\frac{A R^{p - 1}}{A R^{r - 1}})^{q}$

= $R^{(r - q) p + (q - p) r + (p - r) q}$ = $R^{0}$ = 1

Question 30.

Let ${a_{n}}$ be a non - constant arithmetic progression with $a_{1} = 1$ and for any $n \geq 1$ , the value
$\frac{a_{2 n} + a_{2 n - 1} + \dots + a_{n + 1}}{a_{n} + a_{n - 1} + \dots a_{1}}$
remains constant.
Then, $a_{15}$ will be?