Sequences and Series — Page 18 | JEE Mathematics

Q171

Five numbers are in A.P. whose sum is 25 and product is 2520. If one of these five numbers is -

{1 \over 2}

, then the greatest number amongst them is:

A

{{21} \over 2}

B 27

C 7

D 16

Correct Answer

Option D

Solution

Let the A.P is a - 2d, a - d, a, a + d, a + 2d $\because$ sum = 25 $\Rightarrow$ 5a = 25 $\Rightarrow$ a = 5 Also given, product (a2 – 4d2) (a2 – d2).a = 2520 $\Rightarrow$ (25 – 4d2) (25 –d2)5 = 2520 $\Rightarrow$ 4d4 – 121d2 – 4d2 + 121 = 0 $\Rightarrow$ (d2 – 1) (4d2 – 121) = 0 $\Rightarrow$ d = $\pm$ 1, d =

\pm {{11} \over 2}

When d = $\pm$ 1 we can't get any fraction term like -

{1 \over 2}

. $\therefore$ d =

\pm {{11} \over 2}

And when d =

{{11} \over 2}

we get largest term = 5 + 2d = 5 + 11 = 16

Q172

If the 10th term of an A.P. is

{1 \over {20}}

and its 20th term is

{1 \over {10}}

, then the sum of its first 200 terms is

A 100

B

100{1 \over 2}

C

50{1 \over 4}

D 50

Correct Answer

Option B

Solution

T10 = a + 9d =

{1 \over {20}}

....(1) T20 = a + 19d =

{1 \over {10}}

.....(2) Equation (2) – (1) 10d =

{1 \over {10}}

-

{1 \over {20}}

$\Rightarrow$ d =

{1 \over {200}}

a +

{9 \over {200}}

=

{1 \over {20}}

$\Rightarrow$ a =

{1 \over {200}}

S200 =

{{200} \over 2}\left[ {{2 \over {200}} + \left( {200 - 1} \right) \times {1 \over {200}}} \right]

=

100\left[ {{2 \over {200}} + {{199} \over {200}}} \right]

=

{{201} \over 2}

=

100{1 \over 2}

Q173

If

0 < \theta ,\phi < {\pi \over 2},x = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } ,y = \sum\limits_{n = 0}^\infty {{{\sin }^{2n}}\phi }

and

z = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta .{{\sin }^{2n}}\phi }

then :

A xy

-

z = (x + y)z

B xyz = 4

C xy + z = (x + y)z

D xy + yz + zx = z

Correct Answer

Option C

Solution

x = 1 + {\cos ^2}\theta + ..........\infty

x = {1 \over {1 - {{\cos }^2}\theta }} = {1 \over {{{\sin }^2}\theta }}

.......(1)

y = 1 + {\sin ^2}\phi + ........\infty

y = {1 \over {1 - {{\sin }^2}\phi }} = {1 \over {{{\cos }^2}\phi }}

....... (2)

z = {1 \over {1 - {{\cos }^2}\theta .{{\sin }^2}\phi }} = {1 \over {1 - \left( {1 - {1 \over x}} \right)\left( {1 - {1 \over y}} \right)}} = {{xy} \over {xy - (x - 1)(y - 1)}}

$\Rightarrow$

xz + yz - z = xy

$\Rightarrow$

xy + z = (x + y)z

Q174

The minimum value of

f(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}

, where a,

x \in R

and a > 0, is equal to :

A

a + {1 \over a}

B 2a

C a + 1

D

2\sqrt a

Correct Answer

Option D

Solution

We know,

AM \ge GM

$\therefore$

{{{a^{a^x}} + {a \over {{a^{a^x}}}}} \over 2} \ge {\left( {{a^{a^x}}\,.\,{a \over {{a^{a^x}}}}} \right)^{1/2}}

\Rightarrow {a^{a^x}} + {a^{1 - a^x}} \ge 2\sqrt a

Q175

If for x, y

\in

R, x > 0, y = log10x + log10x1/3 + log10x1/9 + ...... upto

\infty

terms and

{{2 + 4 + 6 + .... + 2y} \over {3 + 6 + 9 + ..... + 3y}} = {4 \over {{{\log }_{10}}x}}

, then the ordered pair (x, y) is equal to :

A (106, 6)

B (104, 6)

C (102, 3)

D (106, 9)

Correct Answer

Option D

Solution

{{2(1 + 2 + 3 + .... + y)} \over {3(1 + 2 + 3 + .... + y)}} = {4 \over {{{\log }_{10}}x}}

\Rightarrow {\log _{10}}x = 6 \Rightarrow x = {10^6}

Now,

y = ({\log _{10}}x) + \left( {{{\log }_{10}}{x^{{1 \over 3}}}} \right) + \left( {{{\log }_{10}}{x^{{1 \over 9}}}} \right) + ....\infty

= \left( {1 + {1 \over 3} + {1 \over 9} + ....\infty } \right){\log _{10}}x

= \left( {{1 \over {1 - {1 \over 3}}}} \right){\log _{10}}x = 9

So, (x, y) = (106, 9)

Q176

Let a1, a2, a3, . . . . . . . , an, . . . . . be in A.P. If a3 + a7 + a11 + a15 = 72, then the sum of its first 17 terms is equal to :

A 306

B 153

C 612

D 204

Correct Answer

Option A

Solution

As a1 a2 . . . . . an . . . . . are in A.P. $\therefore$ a3 + a15 = a7 + a11 = a1 + a17 Given, a3 + a7 + a11 + a15 + a15 = 72 $\Rightarrow$ (a3 + a15) + (a7 + a11) = 72 $\Rightarrow$ 2(a1 + a17) = 72 $\Rightarrow$ (a1 + a17) = 36 $\therefore$ Sum of first 17 terms =

{{17} \over 2}

(a1 + a17) =

{{17} \over 2}

$\times$ 36 = 306

Q177

If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is :

A 21

B 22

C 23

D 24

Correct Answer

Option C

Solution

a, A1, A2 ........... An, 100 Let d be the common difference of above A.P. then

{{a + d} \over {100 - d}} = {1 \over 7}

\Rightarrow 7a + 8d = 100

...... (i) and

a + n = 33

..... (ii) and

100 = a + (n + 1)d

\Rightarrow 100 = a + (34 - a){{(100 - 7a)} \over 8}

\Rightarrow 800 = 8a + 7{a^2} - 338a + 3400

\Rightarrow 7{a^2} - 330a + 2600 = 0

\Rightarrow a = 10,\,{{260} \over 7},

but

a \ne {{260} \over 7}

$\therefore$

n = 23

Q178

Let A1, A2, A3, ....... be an increasing geometric progression of positive real numbers. If A1A3A5A7 =

{1 \over {1296}}

and A2 + A4 =

{7 \over {36}}

, then the value of A6 + A8 + A10 is equal to

A 33

B 37

C 43

D 47

Correct Answer

Option C

Solution

{{{A_4}} \over {{r^3}}}.\,{{{A_4}} \over r}.\,{A_4}r\,.\,{A_4}{r^3} = {1 \over {1296}}

{A_4} = {1 \over 6}

{A_2} = {7 \over {36}} - {1 \over 6} = {1 \over {36}}

So

{A_6} + {A_8} + {A_{10}} = 1 + 6 + 36 = 43

Q179

The sum

\sum\limits_{n = 1}^\infty {{{2{n^2} + 3n + 4} \over {(2n)!}}}

is equal to :

A

{{11e} \over 2} + {7 \over {2e}}

B

{{13e} \over 4} + {5 \over {4e}} - 4

C

{{11e} \over 2} + {7 \over {2e}} - 4

D

{{13e} \over 4} + {5 \over {4e}}

Correct Answer

Option B

Solution

$\begin{aligned} & \sum_{n=1}^{\infty} \dfrac{2 n^2+3 n+4}{(2 n) !} \\\\ &= \dfrac{1}{2} \sum_{n=1}^{\infty} \dfrac{2 n(2 n-1)+8 n+8}{(2 n) !} \\\\ &= \dfrac{1}{2} \sum_{n=1}^{\infty} \dfrac{1}{(2 n-2) !}+2 \sum_{n=1}^{\infty} \dfrac{1}{(2 n-1) !}+4 \sum_{n=1}^{\infty} \dfrac{1}{(2 n) !} \\\\ & e=1+1+\dfrac{1}{2 !}+\dfrac{1}{3 !}+\dfrac{1}{4 !}+\ldots . \\\\ & e^{-1}=1-1+\dfrac{1}{2 !}-\dfrac{1}{3 !}+\dfrac{1}{4 !}+\ldots . \\\\ & \left(\mathrm{e}+\dfrac{1}{\mathrm{e}}\right)=2\left(1+\dfrac{1}{2 !}+\dfrac{1}{4 !}+\ldots . .\right) \\\\ & e-\dfrac{1}{e}=\left(1+\dfrac{1}{3 !}+\dfrac{1}{5 !}+\ldots . .\right)\end{aligned}$ Now

\begin{aligned} & \frac{1}{2}\left(\sum_{n=1}^{\infty} \frac{1}{(2 n-2) !}\right)+2 \sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}+4 \sum_{n=1}^{\infty} \frac{1}{(2 n) !} \\\\ & =\frac{1}{2}\left[\frac{e+\frac{1}{\mathrm{e}}}{2}\right]+2\left[\frac{\mathrm{e}-\frac{1}{\mathrm{e}}}{2}\right]+4\left[\frac{\mathrm{e}+\frac{1}{\mathrm{e}}-2}{2}\right] \\\\ & =\frac{\left(\mathrm{e}+\frac{1}{\mathrm{e}}\right)}{4}+e-\frac{1}{\mathrm{e}}+2 \mathrm{e}+\frac{2}{\mathrm{e}}-4 \\\\ & =\frac{13}{4} e+\frac{5}{4 e}-4 \end{aligned}

Q180

If

\mathrm{S}_{n}=4+11+21+34+50+\ldots

to

n

terms, then

\dfrac{1}{60}\left(\mathrm{~S}_{29}-\mathrm{S}_{9}\right)

is equal to :

A 227

B 226

C 220

D 223

Correct Answer

Option D

Solution

Given that

\begin{aligned} & \mathrm{S}_n=4+11+21+24+50+\ldots+\mathrm{T}_n \\\\ & \mathrm{~S}_n=~~~~~~~~4+11+21+34++\mathrm{T}_{n-1}+\mathrm{T}_n \\ & -\quad-\quad-\quad-\quad-\quad-\quad- \\ & \\ 0=4+7+10+13+16+\ldots\left(\mathrm{T}_n-\mathrm{T}_{n-2}\right)-\mathrm{T}_n \end{aligned}

\begin{aligned} & \Rightarrow \mathrm{T}_n=4+7+10+13+16+\ldots . \text{ to } n \text{ terms } \\\\ & \Rightarrow \mathrm{T}_n=\frac{n}{2}[2 \times 4+(n-1) 3] \\\\ & \mathrm{T}_n=\frac{3}{2} n^2+\frac{5}{2} n \end{aligned}

\begin{aligned} & \text{ So } \mathrm{S}_n=\Sigma \mathrm{T}_n=\frac{3}{2} \Sigma n^2+\frac{5}{2} \Sigma n \\\\ & \Rightarrow \mathrm{S}_n=\frac{3}{2} \times \frac{n(n+1)(2 n+1)}{6}+\frac{5}{2} \times \frac{n(n+1)}{2} \end{aligned}

\begin{aligned} & \Rightarrow S_n=\frac{n(n+1)}{4}(2 n+1+5)=\frac{n(n+1)(n+3)}{2} \\\\ & \text{ Hence, } \frac{1}{60}\left(S_{29}-S_9\right)=\frac{1}{60} \times \frac{1}{2}(29 \times 30 \times 32-9 \times 10 \times 12) \\\\ & =223 \end{aligned}