Sequences and Series Questions — JEE Mathematics

Q1

If b is the first term of an infinite G.P. whose sum is five, then b lies in the interval :

A (

-

\infty

,

-

10]

B (

-

10, 0)

C (0, 10)

D [10,

\infty

)

Correct Answer

Option C

Solution

Sum of infinite G.P, S =

b \over {1-r}

where

\left| r \right| < 1

$\Rightarrow$ 5 =

b \over {1-r}

$\Rightarrow$ 1 - r =

b \over 5

$\Rightarrow$ b = 5(1 - r) as

\left| r \right| < 1

$\therefore$ -1 < r < 1 $\Rightarrow$ 1 > -r > -1 $\Rightarrow$ 2 > 1-r > 0 $\Rightarrow$ 10 > 5(1-r) > 0 $\Rightarrow$ 10 > b > 0 $\therefore$ interval of b = (0, 10)

Q2

Let x, y, z be positive real numbers such that x + y + z = 12 and x3y4z5 = (0.1) (600)3. Then x3 + y3 + z3is equal to :

A 270

B 258

C 342

D 216

Correct Answer

Option D

Solution

As we know AM $\ge$ GM $\Rightarrow$

{{3\left( {{x \over 3}} \right) + 4\left( {{y \over 4}} \right) + 5\left( {{z \over 5}} \right)} \over {12}}

$\ge$

{\left[ {{{\left( {{x \over 3}} \right)}^3}{{\left( {{y \over 4}} \right)}^4}{{\left( {{z \over 5}} \right)}^5}} \right]^{{1 \over {12}}}}

$\Rightarrow$ 1 $\ge$

{{{x^3}{y^4}{z^5}} \over {{3^3}{4^4}{5^5}}}

$\Rightarrow$ x3 y4 z5 $\le$ 33 .

44 . 55 $\Rightarrow$ x3 y4 z5 $\le$ (0.1)(600)3 but given that, x3 y4 z5 = (0.1) (600)3 $\therefore$ AM $=$ GM $\Rightarrow$ All the number are equal.

$\therefore$

{x \over 3} = {y \over 4} = {z \over 5} = k

$\Rightarrow$ x $=$ 3k, y = 4k, z = 5k given that, x + y + z $=$ 12 $\Rightarrow$ 3k + 4k + 5k $=$ 12 $\Rightarrow$ 12k $=$ 12 $\Rightarrow$ k = 1 $\therefore$ x $=$ 3, y $=$ 4, z $=$ 5 So, x3 + y3 + z3 $=$ 33 + 43 + 53 $=$ 216

Q3

Let Sk =

{{1 + 2 + 3 + .... + k} \over k}.

If

S_1^2 + S_2^2 + .....\, + S_{10}^2 = {5 \over {12}}

A, then A is equal to :

A 283

B 156

C 301

D 303

Correct Answer

Option D

Solution

Sk =

{{K + 1} \over 2}

\sum {S_k^2} = {5 \over {12}}

A

\sum\limits_{K = 1}^{10} {{{\left( {{{K + 1} \over 2}} \right)}^2}} = {{{2^2} + {3^2} + - - + {{11}^2}} \over 4} = {5 \over {12}}

A

{{11 \times 12 \times 23} \over 6} - 1 = {5 \over 3}

A 505

= {5 \over 3}

A, A = 303

Q4

If

\dfrac{1}{1^4} + \dfrac{1}{2^4} + \dfrac{1}{3^4} + \ldots \infty= \dfrac{\pi^4}{90}

,

\dfrac{1}{1^4} + \dfrac{1}{3^4} + \dfrac{1}{5^4} + \ldots \infty= \alpha

,

\dfrac{1}{2^4} + \dfrac{1}{4^4} + \dfrac{1}{6^4} + \ldots \infty= \beta

, then

\dfrac{\alpha}{\beta}

is equal to :

A 23

B 14

C 18

D 15

Correct Answer

Option D

Solution

\begin{aligned} & \text{ If } \frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots . \infty=\frac{\pi^4}{90} \quad\text{....... (i)}\\ & \beta=\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots, \\ & =\frac{1}{16}\left[\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots . .\right], \end{aligned}

$=\dfrac{1}{16} \times \dfrac{\pi^4}{90}$ using (ii) ............ (ii)

\begin{aligned} & \alpha=\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\ldots . . \infty \\ & \left(\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\ldots . .\right) \\ & -\left(\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots . .\right) \\ & \alpha=\frac{\pi^4}{90}-\frac{1}{16} \times \frac{\pi^4}{90} \quad \text{ [using (i) and (ii)] } \\ & \alpha=\frac{16-1}{16 \times 90} \times \pi^4=\frac{15}{16 \times 90} \pi^4=\frac{\pi^4}{96} \\ & \therefore \frac{\alpha}{\beta}=\frac{\frac{\pi^4}{96}}{\frac{\pi^4}{16 \times 90}}=\frac{16 \times 90}{96}=15 \end{aligned}

Q5

Let

a

and

b

be be two distinct positive real numbers. Let

11^{\text{th }}

term of a GP, whose first term is

a

and third term is

b

, is equal to

p^{\text{th }}

term of another GP, whose first term is

a

and fifth term is

b

. Then

p

is equal to

A 20

B 24

C 21

D 25

Correct Answer

Option C

Solution

The problem involves finding a relation between terms of two different geometric progressions (GPs) which share common first terms but have different terms equated to the same value.

We solve this by setting up equations based on the given conditions for each GP and comparing the terms specified to be equal.For the first GP, with first term

a

and third term

b

:The third term is given by

t_3 = ar^2 = b

, leading to

r^2 = \frac{b}{a}

.The eleventh term is

t_{11} = ar^{10} = a\left(\frac{b}{a}\right)^5

.For the second GP, with first term

a

and fifth term

b

:The fifth term is

T_5 = ar^4 = b

, yielding

r^4 = \frac{b}{a}

and

r = \left(\frac{b}{a}\right)^{1/4}

.The

p^{\text{th}}

term is thus

T_p = ar^{p-1} = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}

.Equating the eleventh term of the first GP to the

p^{\text{th}}

term of the second GP gives:

a\left(\frac{b}{a}\right)^5 = a\left(\frac{b}{a}\right)^{\frac{p-1}{4}}

.Simplifying, we find that

5 = \frac{p-1}{4}

, leading to

p = 21

, which is the solution.

Q6

{1^3} - \,\,{2^3} + {3^3} - {4^3} + ... + {9^3} =

A 425

B - 425

C 475

D - 475

Correct Answer

Option A

Solution

{1^3} - {2^3} + {3^3} - {4^3} + ...... + {9^3}

= {1^3} + {2^3} + {3^3} + ...... + {9^3}

\,\,\,\,\,\,\,\,\,\, - 2\left( {{2^3} + {4^3} + {6^3} + {8^3}} \right)

= {\left[ {{{9 \times 10} \over 2}} \right]^2} - {2.2^3}\left[ {{1^3} + {2^3} + {3^3} + {4^3}} \right]

= {\left( {45} \right)^2} - 16.{\left[ {{{4 \times 5} \over 2}} \right]^2}

= 2025 - 1600 = 425

Q7

The sum of the serier

{1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..............

up to

\infty

is equal to

A

\log {\,_e}\left( {{4 \over e}} \right)\,\,

B

2\,\log {\,_e}2

C

\log {\,_e}2 - 1\,

D

\log {\,_e}2

Correct Answer

Option A

Solution

{1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..........\infty

\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)

S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty

= \left( {{1 \over 1} - {1 \over 2}} \right) - \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right)

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{1 \over 4} - {1 \over 5}} \right).......

= 1 - 2\left[ {{1 \over 2} - {1 \over 3} + {1 \over 4} - {1 \over 5}.........\infty } \right]

= 1 - 2\left[ { - \log \left( {1 + 1} \right) + 1} \right]

= 2\log 2 - 1 = \log \left( {{4 \over e}} \right)

Q8

The sum of series

{1 \over {2\,!}} + {1 \over {4\,!}} + {1 \over {6\,!}} + ........

is

A

{{\left( {{e^2} - 2} \right)} \over e}\,

B

{{{{\left( {e - 1} \right)}^2}} \over {2e}}

C

{{\left( {{e^2} - 1} \right)} \over {2e}}\,

D

{{\left( {{e^2} - 1} \right)} \over 2}

Correct Answer

Option B

Solution

We know that

e = 1 + {1 \over {1!}} + {1 \over {2!}} + {1 \over {3!}} + ..........

and

{e^{ - 1}} = 1 - {1 \over {1!}} + {1 \over {2!}} - {1 \over {3!}} + .........

$\therefore$

e + {e^{ - 1}} = 2\left[ {1 + {1 \over {2!}} + {1 \over {4!}} + ......} \right]

$\therefore$

{1 \over {2!}} + {1 \over {4!}} + {1 \over {6!}} + .......

= {{e + {e^{ - 1}}} \over 2} - 1

= {{{e^2} + 1 - 2e} \over {2e}}

= {{{{\left( {e - 1} \right)}^2}} \over {2e}}

Q9

Let

{{T_r}}

be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n,

m \ne n,\,\,{T_m} = {1 \over n}\,\,and\,{T_n} = {1 \over m},\,

then a - d equals

A

{1 \over m} + {1 \over n}

B 1

C

{1 \over {m\,n}}

D 0

Correct Answer

Option D

Solution

{T_m} = a + \left( {m - 1} \right)d = {1 \over n}...........\left( 1 \right)

{T_n} = a + \left( {n - 1} \right)d = {1 \over m}..........\left( 2 \right)

\left( 1 \right) - \left( 2 \right) \Rightarrow \left( {m - n} \right)d

= {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn}}

From

\left( 1 \right)

a = {1 \over {mn}} \Rightarrow a - d = 0

Q10

The sum of the first n terms of the series

{1^2} + {2.2^2} + {3^2} + {2.4^2} + {5^2} + {2.6^2} + ....\,is\,{{n{{(n + 1)}^2}} \over 2}

when n is even. When n is odd the sum is

A

{\left[ {{{n(n + 1)} \over 2}} \right]^2}

B

{{{n^2}(n + 1)} \over 2}

C

{{n{{(n + 1)}^2}} \over 4}

D

\,{{3n(n + 1)} \over 2}

Correct Answer

Option B

Solution

If

n

is odd, the required sum is

{1^2} + {2.2^2} + {3^2} + {2.4^2} + ......

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}

= {{\left( {n - 1} \right){{\left( {n - 1 + 1} \right)}^2}} \over 2} + {n^2}

[ As

\left( {n - 1} \right)

is even $\therefore$ using given formula for the sum of

\left( {n - 1} \right)

terms.]

= \left( {{{n - 1} \over 2} + 1} \right){n^2} = {{{n^2}\left( {n + 1} \right)} \over 2}