Sequences And Series(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

SEQUENCES AND SERIES MCQs

Sequence And Series

Total Questions : 75 | Page 2 of 8 pages

Question 11. The sum of infinite terms of the following series
1 +

\frac{4}{5}

\frac{7}{5^{2}}

\frac{10}{5^{3}}

+ ...........∞ will be

316
358
354
3516

Discuss Question

Answer: Option D. -> 3516
:
D
Let the sum to infinity of thearithmetic-geometric series be
S = 1 + 4 .

\frac{1}{5}

+ 7.

\frac{1}{5^{2}}

+ 10.

\frac{1}{5^{3}}

+........
⇒

\frac{1}{5}

S =

\frac{1}{5}

+ 4.

\frac{1}{5^{2}}

+ 7.

\frac{1}{5^{3}}

+...........
Subtracting (1 -

\frac{1}{5}

)S = 1 + 3.

\frac{1}{5}

+ 3.

\frac{1}{5^{2}}

+ 3.

\frac{1}{5^{3}}

+ .......
= 1 + 3(

\frac{1}{5}

\frac{1}{5^{2}}

+ ..............)
⇒

\frac{4}{5}

S = 1 + 3.

\frac{1}{5}

(

\frac{1}{1 - \frac{1}{5}}

) = 1 +

\frac{3}{4}

\frac{7}{4}

⇒ S =

\frac{35}{16}

.
Aliter :Use direct formula

S_{\infty}

\frac{a b}{1 - r}

\frac{d b r}{(1 - r)^{2}}

Here a = 1, b = 1, d = 3, r =

\frac{1}{5}

, therefore

S_{\infty}

\frac{1}{1 - \frac{1}{5}}

\frac{3 \times 1 \times \frac{1}{5}}{(1 - {\frac{1}{5}}^{2})}

\frac{5}{4}

\frac{\frac{3}{5}}{\frac{16}{25}}

\frac{5}{4}

\frac{15}{16}

\frac{35}{16}

.
Aliter:Use S = [1 +

\frac{r}{1 - r}

× diff. of A.P.]

\frac{1}{1 - r}

Question 12. Harikiran purchased a house in Rs. 15000 and paid Rs. 5000 at once. Rest money he promised to pay in annual instalment of Rs. 1000 with 10% per annum interest. How much money is to be paid by him

Rs. 21555
Rs. 20475
Rs. 20500
Rs. 20700

Discuss Question

Answer: Option C. -> Rs. 20500
:
C
It will take 10 years for harikiranto pay off Rs. 10000 in 10 yearly
installments.
∵ He pays 10% annual interest on interest on remaining amount
∴ Money given in first year
= 1000 +

\frac{10000 \times 10}{100}

= Rs. 2000
Money given in second year = 1000 + interest of (10000 - 1000) with interest rate 10% per annum
= 1000 +

\frac{9000 \times 10}{100}

= Rs. 1900
Money paid in third year = Rs. 1800 etc.
So money given by Jairam in 10 years will be Rs. 2000, Rs.1900, Rs. 1800, Rs. 1700,......,
which is in arithmetic progression,
whose first term a = 2000 and d = -100
Total money given in 10 years = sum of 10 terms of arithmetic
progression
=

\frac{10}{2}

[2(2000) + (10 - 1)(-100)] = Rs. 15500
Therefore, total money given by harikiran
= 5000 + 15500 = Rs. 20500.

Question 13. The sum of the first n terms is

\frac{1}{2}

\frac{3}{4}

\frac{7}{8}

\frac{15}{16}

+ .......... is

2n - n - 1
1 - 2−n
n + 2−n - 1
2n - 1

Discuss Question

Answer: Option C. -> n + 2−n - 1
:
C
The sum of the first n terms is

S_{n}

= (1 -

\frac{1}{2}

) + (1 -

\frac{1}{2^{2}}

) + (1 -

\frac{1}{2^{3}}

) + (1 -

\frac{1}{2^{4}}

+ ........... + (1 -

\frac{1}{2^{n}}

)
= n - {

\frac{1}{2}

\frac{1}{2^{2}}

+............+

\frac{1}{2^{n}}

}
= n -

\frac{1}{2}

(

\frac{1 - \frac{1}{2^{n}}}{1 - \frac{1}{2}}

) = n - (1 -

\frac{1}{2^{n}}

) = n - 1 +

2^{- n}

.
Trick:Check for n = 1, 2 i.e.,

S_{1}

\frac{1}{2}

S_{2}

\frac{5}{4}

and
(c) ⇒

S_{1}

\frac{1}{2}

and

S_{2}

= 2 +

2^{- 2}

- 1 =

\frac{5}{4}

Question 14. If the aritmetic, geometric and harmonic menas between two positive real numbers be A, G and H, then

A2 = GH
H2 = AG
G = AH
G2 = AH

Discuss Question

Answer: Option D. -> G2 = AH
:
D
Let A =

\frac{a + b}{2}

, G =

\sqrt{a b}

and H =

\frac{2 a b}{a + b}

.
Then

G^{2}

= ab ..................(i)
and AH =

(\frac{a + b}{2}) . \frac{2 a b}{a + b}

= ab ................(ii)
From (i) and (ii), we have

G^{2}

= AH

Question 15. If p, q, r are in A.P. and are positive, the roots of the quadratic
equation p

x^{2}

+ qx + r = 0 are all real for ___.

| rp - 7| ≥ 4 √3
| pr - 7|
All p and r
No p and r

Discuss Question

Answer: Option A. -> | rp - 7| ≥ 4 √3
:
A
p,q,r are positive and are in A.P.
∴ q =

\frac{p + r}{2}

.........(i)
∵ The roots ofp

x^{2}

+ qx + r = 0 are real
⇒

q^{2}

≥ 4pr⇒

{[\frac{p + r}{2}]}^{2}

≥ 4pr [using (i)]
⇒

p^{2}

r^{2}

- 14pr≥ 0
⇒

{(\frac{r}{p})}^{2}

- 14 (

\frac{r}{p}

) + 1≥ 0 ( ∵ p>0 and p≠ 0)

{(\frac{r}{p} - 7)}^{2}

({4 \sqrt{3})}^{2}

≥ 0
⇒|

\frac{r}{p}

- 7|≥ 4

\sqrt{3}

Question 16. 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 ___.

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 17. 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?

30
29
31
Can't be determined

Discuss Question

Answer: Option B. -> 29
:
B

C = \frac{a_{2 n} + a_{2 n - 1} + a_{2 n - 2} + \dots + a_{n + 1}}{a_{n} + a_{n - 1} + \dots + a_{1}}

Adding 1 on both sides, we get

C + 1 = \frac{a_{2 n} + a_{2 n - 1} + a_{2 n - 2} + \dots + a_{n + 1}}{a_{n} + a_{n - 1} + \dots + a_{1}}

= \frac{a_{2 n} + a_{2 n - 1} + \dots + a_{n + 1} + a_{n} + \dots + a_{1}}{a_{n} + a_{n - 1} + \dots + a_{1}} ∴ C + 1 = \frac{S_{2 n}}{S_{n}} = \frac{\frac{2 n}{2} [2 a_{1} + (2 n - 1) d]}{\frac{n}{2} [2 a_{1} + (n - 1) d]} = \frac{2 [2 a_{1} + (2 n - 1) d]}{2 a_{1} + (n - 1) d}

Since,

\frac{C + 1}{2}

is a constant.
Let

\frac{C + 1}{2} = R

\Rightarrow \frac{2 + (2 n - 1) d}{2 + (n - 1) d} = R [∵ a_{1} = 1] \Rightarrow 2 + (2 n - 1) d = 2 R + (n - 1) d R \Rightarrow 2 R + n d R - d R = 2 + 2 n d - d \Rightarrow n d (R - 2) = d R - 2 R - d + 2 \Rightarrow n d (R - 2) = (d - 2) (R - 1)

Now, left side has n which changes and right side remains constant

∴

LHS = RHS = 0

\Rightarrow R = 2 [∵ n \neq 0, d \neq 0] \Rightarrow 0 = d - 2 \Rightarrow d = 2 ∴ a_{15} = a_{1} + (15 - 1) d = 1 + 14 \times 2 = 29

Question 18.

2^{\frac{1}{4}}

4^{\frac{1}{8}}

8^{\frac{1}{16}}

16^{\frac{1}{32}}

................ is equal to

Discuss Question

Answer: Option B. -> 2
:
B

2^{\frac{1}{4}}

4^{\frac{1}{8}}

8^{\frac{1}{16}}

16^{\frac{1}{32}}

.........∞
=

2^{\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + . . . . . . . .}

2^{S}

, where S is given by
S =

\frac{1}{4}

+ 2

\frac{1}{8}

+ 3

\frac{1}{16}

\frac{1}{32}

+ ............∞ .........(i)
⇒

\frac{1}{2}

S =

\frac{1}{8}

\frac{2}{16}

\frac{3}{32}

\frac{4}{64}

+ ................∞ ..........(ii)
Subtracting (ii) from (i), we get S = 1.
Hence required product =

2^{1}

= 2.

Question 19. If the aritmetic, geometric and harmonic menas between two positive real numbers be A, G and H, then

A2 = GH
H2 = AG
G = AH
G2 = AH

Discuss Question

Answer: Option D. -> G2 = AH
:
D
Let A =

\frac{a + b}{2}

, G =

\sqrt{a b}

and H =

\frac{2 a b}{a + b}

.
Then

G^{2}

= ab ..................(i)
and AH =

(\frac{a + b}{2}) . \frac{2 a b}{a + b}

= ab ................(ii)
From (i) and (ii), we have

G^{2}

= AH

Question 20. The sum of the first n terms is

\frac{1}{2}

\frac{3}{4}

\frac{7}{8}

\frac{15}{16}

+ .......... is

2n - n - 1
1 - 2−n
n + 2−n - 1
2n - 1

Discuss Question

Answer: Option C. -> n + 2−n - 1
:
C
The sum of the first n terms is

S_{n}

= (1 -

\frac{1}{2}

) + (1 -

\frac{1}{2^{2}}

) + (1 -

\frac{1}{2^{3}}

) + (1 -

\frac{1}{2^{4}}

+ ........... + (1 -

\frac{1}{2^{n}}

)
= n - {

\frac{1}{2}

\frac{1}{2^{2}}

+............+

\frac{1}{2^{n}}

}
= n -

\frac{1}{2}

(

\frac{1 - \frac{1}{2^{n}}}{1 - \frac{1}{2}}

) = n - (1 -

\frac{1}{2^{n}}

) = n - 1 +

2^{- n}

.
Trick:Check for n = 1, 2 i.e.,

S_{1}

\frac{1}{2}

S_{2}

\frac{5}{4}

and
(c) ⇒

S_{1}

\frac{1}{2}

and

S_{2}

= 2 +

2^{- 2}

- 1 =

\frac{5}{4}

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