Sequences and Series — Page 20 | JEE Mathematics

Q191

The sum of the series

1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......

ad inf. is

A

{{e - 1} \over {\sqrt e }}\,

B

{{e + 1} \over {\sqrt e }}

C

{{e - 1} \over {2\sqrt e }}

D

{{e + 1} \over {2\sqrt e }}

Correct Answer

Option D

Solution

{{{e^x} + {e^{ - x}}} \over 2}

= 1 + {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} + {{{x^6}} \over {6!}}.........

Putting

x = {1 \over 2}

we get

1 + {1 \over {4.2!}} + {1 \over {16.4!}} + {1 \over {64.6!}} + .......

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

= {{e + 1} \over {2\sqrt e }}

Q192

If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is :

A 2

B 4

{^{{1 \over 3}}}

C 4

{^{{2 \over 3}}}

D 4

Correct Answer

Option A

Solution

a, b and c are in AP. $\therefore$ a + c = 2b As, abc = 8 $\Rightarrow$ ac

\left( {{{a + c} \over 2}} \right)

= 8 $\Rightarrow$ ac(a + c) = 16 = 4 $\times$ 4 $\therefore$ ac = 4 and a + c = 4 Then, b =

\left( {{{a + c} \over 2}} \right)

=

{4 \over 2}

= 2

Q193

The value of

\,{2^{1/4}}.\,\,{4^{1/8}}.\,{8^{1/16}}...\infty

is

A 1

B 2

C 3/2

D 4

Correct Answer

Option B

Solution

The product is

p = {2^{1/4}}{.2^{2/8}}{.2^{3/16}}........

= {2^{1/4 + 2/8 + 3/16 + .......\infty }}

Now let

S = {1 \over 4} + {2 \over 8} + {3 \over {16}} + .......\infty \,\,\,\,........\left( 1 \right)

{1 \over 2}S = {1 \over 8} + {2 \over {16}} + .......\infty \,\,\,\,........\left( 2 \right)

Subtracting

(2)

from

(1)

\Rightarrow {1 \over 2}S = {1 \over 4} + {1 \over 8} + {1 \over {16}} + .......\infty

or

{1 \over 2}S = {{1/4} \over {1 - 1/2}} = {1 \over 2} \Rightarrow S = 1

$\therefore$

P = {2^S} = 2

Q194

If

x = \sum\limits_{n = 0}^\infty {{a^n},\,\,y = \sum\limits_{n = 0}^\infty {{b^n},\,\,z = \sum\limits_{n = 0}^\infty {{c^n},} } } \,\,

where a, b, c are in A.P and

\,\left| a \right| < 1,\,\left| b \right| < 1,\,\left| c \right| < 1

then x, y, z are in

A G.P.

B A.P.

C Arithmetic-Geometric Progression

D H.P.

Correct Answer

Option D

Solution

x = \sum\limits_{n = 0}^\infty {{a^n}} = {1 \over {1 - a}}\,\,\,\,\,\,\,\,\,\,a = 1 - {1 \over x}

y = \sum\limits_{n = 0}^\infty {{b^n}} = {1 \over {1 - b}}\,\,\,\,\,\,\,\,\,\,b = 1 - {1 \over y}

z = \sum\limits_{n = 0}^\infty {{c^n}} = {1 \over {1 - c}}\,\,\,\,\,\,\,\,\,\,c = 1 - {1 \over z}

a,b,c

are in

A.P.

OR

2b = a + c

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

{2 \over y} = {1 \over x} + {1 \over z} \Rightarrow x,y,z

are in

H.P.

Q195

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

A

{2 \over {13}}\left( {{3^{50}} - 1} \right)

B

{1 \over {13}}\left( {{3^{50}} - 1} \right)

C

{1 \over {26}}\left( {{3^{49}} - 1} \right)

D

{1 \over {26}}\left( {{3^{50}} - 1} \right)

Correct Answer

Option D

Solution

Let first term = a > 0 Common ratio = r > 0 ar + ar2 + ar3 = 3 ....(i) ar5 + ar6 + ar7 = 243 ....(ii) $\Rightarrow$ r4(ar + ar2 + ar3) = 243 $\Rightarrow$ r4(3) = 243 $\Rightarrow$ r = 3 as r > 0 from (i) 3a + 9a + 27a = 3 $\Rightarrow$ a =

{1 \over {13}}

$\therefore$ S50 =

{{a\left( {{r^{50}} - 1} \right)} \over {\left( {r - 1} \right)}}

=

{1 \over {26}}\left( {{3^{50}} - 1} \right)

Q196

The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 +.... upto 11th term is :-

A 945

B 916

C 915

D 946

Correct Answer

Option D

Solution

S = 1 + 2 × 3 + 3 × 5 + 4 × 7 +.... upto 11th term General term Tn = n (2n - 1) $\therefore$ Sn =

\sum {{T_n}}

$\Rightarrow$ Sn =

\sum {\left( {2{n^2} - n} \right)}

$\Rightarrow$ Sn =

2\sum {{n^2} - \sum r }

$\Rightarrow$ Sn =

2\left( {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}} \right) - {{n\left( {n + 1} \right)} \over 2}

$\therefore$ S11 =

2\left( {{{11\left( {12} \right)\left( {23} \right)} \over 6}} \right) - {{11\left( {12} \right)} \over 2}

$\Rightarrow$ S11 = 946

Q197

Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression

{{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}}

is :

A

{1 \over 2}

B

{1 \over 4}

C

{{m + n} \over {6mn}}

D 1

Correct Answer

Option B

Solution

{{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}} = {1 \over {\left( {{x^m} + {1 \over {{x^m}}}} \right)\left( {{y^n} + {1 \over {{y^n}}}} \right)}} \le {1 \over 4}

using AM $\ge$ GM

Q198

If nC4, nC5 and nC6 are in A.P., then n can be :

A 11

B 12

C 9

D 14

Correct Answer

Option D

Solution

2.nC5 = nC4 + nC6 2.

{n \over {\left| 5 \right|n - 5}} = {n \over {\left| 4 \right|n - 4}} + {n \over {\left| 6 \right|n - 6}}

{2 \over 5}.{1 \over {n - 5}} = {1 \over {\left( {n - 4} \right)\left( {n - 5} \right)}} + {1 \over {30}}

n = 14

satisfying equation.

Q199

If

\alpha

,

\beta

are natural numbers such that 100

\alpha

-

199

\beta

= (100)(100) + (99)(101) + (98)(102) + ...... + (1)(199), then the slope of the line passing through (

\alpha

,

\beta

) and origin is :

A 540

B 550

C 530

D 510

Correct Answer

Option B

Solution

RHS =

\sum\limits_{r = 0}^{99} {(100 - r)(100 + r)}

= {(100)^3} - {{99 \times 100 \times 199} \over 6} = {(100)^3} - (1650)199

LHS = (100) $\alpha$ $-$ (199) $\beta$ So, $\alpha$ = 3, $\beta$ = 1650 Slope = tan $\theta$ =

{\beta \over \alpha }

$\Rightarrow$ tan $\theta$ = 550

Q200

Let a, b and c be the 7th, 11th and 13th terms respectively of a non-constant A.P. If these are also three consecutive terms of a G.P., then

{a \over c}

equal to :

A 2

B

{1 \over 2}

C

{7 \over 13}

D 4

Correct Answer

Option D

Solution

T7 = A + 6d = a; T11 = A + 10d = b; T13 = A + 12d = c Now a, b, c are in G.P. $\therefore$ b2 = ac $\Rightarrow$ (A + 10d)2 = (A + 6d) (A + 12d) $\Rightarrow$ A2 + 100d2 + 20Ad = A2 + 18Ad + 72d2 $\Rightarrow$ A + 14d = 0, A = $-$ 14d

{a \over c} = {{A + 6d} \over {A + 12d}} = {{ - 8d} \over { - 2d}} = 4