Sequences and Series — Page 4 | JEE Mathematics

Q31

Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If (S2

-

S1) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :

A 7000

B 1000

C 3000

D 5000

Correct Answer

Option C

Solution

S1 =

{{2n} \over 2}

[2a + (2n $-$ 1)d] S2 =

{{4n} \over 2}

[2a + (4n $-$ 1)d] (where a = T1 and d is common difference) S2 $-$ S1 $\Rightarrow$ 2n[2a + (4n $-$ 1)d] $-$ n[2a + (2n $-$ 1)d] = 1000 $\Rightarrow$ n[2a + d(8n $-$ 2 $-$ 2n + 1)] = 1000 $\Rightarrow$ n[2a + (6n $-$ 1)d] = 1000 S6 =

{{6n} \over 2}

[2a + (6n $-$ 1)d] = 3(S2 $-$ S1) = 3000

Q32

Let Sn be the sum of the first n terms of an arithmetic progression. If S3n = 3S2n, then the value of

{{{S_{4n}}} \over {{S_{2n}}}}

is :

A 6

B 4

C 2

D 8

Correct Answer

Option A

Solution

Let a be first term and d be common diff. of this A.P. Given, S3n = 3S2n

\Rightarrow {{3n} \over 2}[2a + (3n - 1)d] = 3{{2n} \over 2}[2a + (2n - 1)d]

\Rightarrow 2a + (3n - 1)d = 4a + (4n - 2)d

\Rightarrow 2a + (n - 1)d = 0

Now,

{{{S_{4n}}} \over {{S_{2n}}}} = {{{{4n} \over 2}[2a + (4n - 1)d]} \over {{{2n} \over 2}[2a + (2n - 1)d]}} = {{2\left[ {\underbrace {2a + (n - 1)d}_{ = 0} + 3nd} \right]} \over {\left[ {\underbrace {2a + (n - 1)d}_{ = 0} + nd} \right]}}

= {{6nd} \over {nd}} = 6

Q33

If sum of the first 21 terms of the series

{\log _{{9^{1/2}}}}x + {\log _{{9^{1/3}}}}x + {\log _{{9^{1/4}}}}x + .......

, where x > 0 is 504, then x is equal to

A 243

B 9

C 7

D 81

Correct Answer

Option D

Solution

s = 2{\log _9}x + 3{\log _9}x + ....... + 22{\log _9}x

s = {\log _9}x(2 + 3 + ..... + 22)

s = {\log _9}x\left\{ {{{21} \over 2}(2 + 22)} \right\}

Given,

252{\log _9}x = 504

\Rightarrow {\log _9}x = 2 \Rightarrow x = 81

Q34

Let Sn denote the sum of first n-terms of an arithmetic progression. If S10 = 530, S5 = 140, then S20

-

S6 is equal to:

A 1862

B 1842

C 1852

D 1872

Correct Answer

Option A

Solution

Let first term of A.P. be a and common difference is d. $\therefore$

{S_{10}} = {{10} \over 2}\{ 2a + 9d\} = 530

$\therefore$

2a + 9d = 106

..... (i)

{S_5} = {5 \over 2}\{ 2a + 4d\} = 140

a + 2d = 28

...... (ii) From equation (i) and (ii), a = 8, d = 10 $\therefore$

{S_{20}} - {S_6} = {{20} \over 2}\{ 2 \times 8 + 19 \times 10\} - {6 \over 2}\{ 2 \times 8 + 5 \times 10\}

= 2060 - 198

= 1862

Q35

The sum of 10 terms of the series

{3 \over {{1^2} \times {2^2}}} + {5 \over {{2^2} \times {3^2}}} + {7 \over {{3^2} \times {4^2}}} + ....

is :

A 1

B

{{120} \over {121}}

C

{{99} \over {100}}

D

{{143} \over {144}}

Correct Answer

Option B

Solution

S = {{{2^2} - {1^2}} \over {{1^2} \times {2^2}}} + {{{3^2} - {2^2}} \over {{2^2} \times {3^2}}} + {{{4^2} - {3^2}} \over {{3^2} \times {4^2}}} + ...

= \left[ {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right] + \left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right] + \left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right] + .... + \left[ {{1 \over {{{10}^2}}} - {1 \over {{{11}^2}}}} \right]

= 1 - {1 \over {121}}

= {{120} \over {121}}

Q36

Let a1, a2, ..........., a21 be an AP such that

\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = {4 \over 9}}

. If the sum of this AP is 189, then a6a16 is equal to :

A 57

B 72

C 48

D 36

Correct Answer

Option B

Solution

\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = \sum\limits_{n = 1}^{20} {{1 \over {{a_n}({a_n} + d)}}} }

= {1 \over d}\sum\limits_{n = 1}^{20} {\left( {{1 \over {{a_n}}} - {1 \over {{a_n} + d}}} \right)}

\Rightarrow {1 \over d}\left( {{1 \over {{a_1}}} - {1 \over {{a_{21}}}}} \right) = {4 \over 9}

(Given)

\Rightarrow {1 \over d}\left( {{{{a_{21}} - {a_1}} \over {{a_1}{a_{21}}}}} \right) = {4 \over 9}

\Rightarrow {1 \over d}\left( {{{{a_1} + 20d - {a_1}} \over {{a_1}{a_2}}}} \right) = {4 \over 9} \Rightarrow {a_1}{a_2} = 45

.... (1) Now sum of first 21 terms =

{{21} \over 2}(2{a_1} + 20d) = 189

$\Rightarrow$ a1 + 10d = 9 ..... (2) For equation (1) & (2) we get a1 = 3 & d =

{3 \over 5}

or a1 = 15 & d =

- {3 \over 5}

So, a6 . a16 = (a1 + 5d) (a1 + 15d) $\Rightarrow$ a6a16 = 72 Option (b)

Q37

If 0 < x < 1 and

y = {1 \over 2}{x^2} + {2 \over 3}{x^3} + {3 \over 4}{x^4} + ....

, then the value of e1 + y at

x = {1 \over 2}

is :

A

{1 \over 2}{e^2}

B 2e

C

{1 \over 2}\sqrt e

D 2e2

Correct Answer

Option A

Solution

y = \left( {1 - {1 \over 2}} \right){x^2} + \left( {1 - {1 \over 3}} \right){x^3} + ....

= ({x^2} + {x^3} + {x^4} + ......) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + ....} \right)

= {{{x^2}} \over {1 - x}} + x - \left( {x + {{{x^2}} \over 2} + {{{x^3}} \over 3} + ....} \right)

= {x \over {1 - x}} + \ln (1 - x)

x = {1 \over 2} \Rightarrow y = 1 - \ln 2

{e^{1 + y}} = {e^{1 + 1 - \ln 2}}

= {e^{2 - \ln 2}} = {{{e^2}} \over 2}

Q38

Let a1, a2, a3, ..... be an A.P. If

{{{a_1} + {a_2} + .... + {a_{10}}} \over {{a_1} + {a_2} + .... + {a_p}}} = {{100} \over {{p^2}}}

, p

\ne

10, then

{{{a_{11}}} \over {{a_{10}}}}

is equal to :

A

{{19} \over {21}}

B

{{100} \over {121}}

C

{{21} \over {19}}

D

{{121} \over {100}}

Correct Answer

Option C

Solution

{{{{10} \over 2}(2{a_1} + 9d)} \over {{p \over 2}(2{a_1} + (p - 1)d)}} = {{100} \over {{p^2}}}

(2{a_1} + 9d)p = 10(2{a_1} + (p - 1)d)

9dp = 20{a_1} - 2p{a_1} + 10d(p - 1)

9p = (20 - 2p){{{a_1}} \over d} + 10(p - 1)

{{{a_1}} \over d} = {{(10 - p)} \over {2(10 - p)}} = {1 \over 2}

$\therefore$

{{{a_{11}}} \over {{a_{10}}}} = {{{a_1} + 10d} \over {{a_1} + 9d}} = {{{1 \over 2} + 10} \over {{1 \over 2} + 9}} = {{21} \over {19}}

Q39

If 0 < x < 1, then

{3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + .....

, is equal to :

A

x\left( {{{1 + x} \over {1 - x}}} \right) + {\log _e}(1 - x)

B

x\left( {{{1 - x} \over {1 + x}}} \right) + {\log _e}(1 - x)

C

{{1 - x} \over {1 + x}} + {\log _e}(1 - x)

D

{{1 + x} \over {1 - x}} + {\log _e}(1 - x)

Correct Answer

Option A

Solution

Let

t = {3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + ......\infty

= \left( {2 - {1 \over 2}} \right){x^2} + \left( {2 - {1 \over 3}} \right){x^3} + \left( {2 - {1 \over 4}} \right){x^4} + ......\infty

= 2({x^2} + {x^3} + {x^4} + .....\infty ) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + .....\infty } \right)

= {{2{x^2}} \over {1 - x}} - (\ln (1 - x) - x)

\Rightarrow t = {{2{x^2}} \over {1 - x}} + x - \ln (1 - x)

\Rightarrow t = {{x(1 + x)} \over {1 - x}} - \ln (1 - x)

Q40

If a1, a2, a3 ...... and b1, b2, b3 ....... are A.P., and a1 = 2, a10 = 3, a1b1 = 1 = a10b10, then a4 b4 is equal to -

A

{{35} \over {27}}

B 1

C

{{27} \over {28}}

D

{{28} \over {27}}

Correct Answer

Option D

Solution

a1, a2, a3 .... are in A.P.

(Let common difference is d1) b1, b2, b3 .... are in A.P.

(Let common difference is d2) and a1 = 2, a10 = 3, a1b1 = 1 = a10b10 $\because$ a1b1 = 1 $\therefore$ b1 =

{1 \over 2}

a10b10 = 1 $\therefore$ b10 =

{1 \over 3}

Now, a10 = a1 + 9d1 $\Rightarrow$ d1 =

{1 \over 9}

b10 = b1 + 9d2 $\Rightarrow$ d2 =

{1 \over 9}

\left[ {{1 \over 3} - {1 \over 2}} \right]

= $-$

{{1 \over {54}}}

Now, a4 = 2 +

{{3 \over {9}}}

=

{{7 \over {3}}}

b4 =

{{1 \over {2}}}

-%

{{3 \over {54}}}

=

{{4 \over {9}}}

$\therefore$ a4b4 =

{{28 \over {27}}}