Sequences and Series — Page 16 | JEE Mathematics

Q151

Fifth term of a GP is 2, then the product of its 9 terms is

A 256

B 512

C 1024

D none of these

Correct Answer

Option B

Solution

a{r^4} = 2

a \times ar \times a{r^2} \times a{r^3} \times a{r^4} \times a{r^5} \times a{r^6} \times a{r^7} \times a{r^8}

= {a^9}{r^{36}} = {\left( {a{r^4}} \right)^9} = {2^9} = 512

Q152

If 1,

{\log _9}\,\,({3^{1 - x}} + 2),\,\,{\log _3}\,\,({4.3^x} - 1)

are in A.P. then x equals

A

{\log _3}\,4\,\,\,

B

1 - \,{\log _3}\,4\,

C

1 - \,{\log _4}\,3

D

{\log _4}\,3

Correct Answer

Option B

Solution

1,\,{\log _9}\left( {{3^{1 - x}} + 2} \right),{\log _3}\left( {{{4.3}^x} - 1} \right)

are in

A.P.

\Rightarrow 2{\log _9}\left( {{3^{1 - x}} + 2} \right)

\,\,\,\,\,\,\,\,\,

= 1 + {\log _3}\left( {{{4.3}^x} - 1} \right)

\Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)

\,\,\,\,\,\,\,\,\,

= {\log _3}3 + {\log _3}\left( {{{4.3}^x} - 1} \right)

\Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)

\,\,\,\,\,\,\,\,\,

= {\log _3}\left[ {3\left( {{{4.3}^x} - 1} \right)} \right]

\Rightarrow {3^{1 - x}} + 2 = 3\,\left( {{{4.3}^x} - 1} \right)

\Rightarrow {3.3^{ - x}} + 2 = {12.3^x} - 3.

Put

{3^x} = t

\Rightarrow {3 \over t} + 2 = 12t - 3

or

12{t^2} - 5t - 3 = 0;

Hence

t = - {1 \over 3},{3 \over 4} \Rightarrow {3^x} = {3 \over 4}

(as

{3^x}\,\, \ne \,\, - ve

)

\Rightarrow x = {\log _3}\left( {{3 \over 4}} \right)

or

x = {\log _3}3 - {\log _3}4

\Rightarrow x = 1 - {\log _3}4

Q153

If the

{2^{nd}},{5^{th}}\,and\,{9^{th}}

terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is :

A 1

B

{7 \over 4}

C

{8 \over 5}

D

{4 \over 3}

Correct Answer

Option D

Solution

The terms of an Arithmetic Progression (A.P.) are given by $a$ , $a + d$ , $a + 2d$ , ..., where $a$ is the first term and $d$ is the common difference.

Given that the 2nd, 5th and 9th terms of an A.P. are in Geometric Progression (G.P.), we can denote them as follows : 2nd term = $a + d$ 5th term = $a + 4d$ 9th term = $a + 8d$ For three numbers to be in G.P., the square of the middle term must be equal to the product of the other two terms.

So, $(a + 4d)^2 = (a + d)(a + 8d)$ Expanding and simplifying : $a^2 + 8ad + 16d^2 = a^2 + 9ad + 8d^2$ $8ad + 16d^2 = a^2 + 9ad + 8d^2$ $8ad - 9ad = 8d^2 - 16d^2$ $-ad = -8d^2$ $a = 8d$ The common ratio of the G.P. is the ratio of the 5th term to the 2nd term, or $(a + 4d) / (a + d)$ .

Substituting $a = 8d$ gives : $(8d + 4d) / (8d + d) = 12d / 9d = 4 / 3$ So, the common ratio of the G.P. is $4 / 3$ .

The answer is option D.

Q154

If m is the A.M. of two distinct real numbers l and n

(l,n > 1)

and

{G_1},{G_2}

and

{G_3}

are three geometric means between

l

and n, then

G_1^4\, + 2G_2^4\, + G_3^4

equals:

A

4\,lm{n^2}

B

4\,{l^2}{m^2}{n^2}

C

4\,{l^2}m\,n

D

4\,l\,{m^2}n

Correct Answer

Option D

Solution

m = {{l + n} \over 2}

and common ratio of

G.P.

= r = {\left( {{n \over l}} \right)^{{1 \over 4}}}

$\therefore$

{G_1} = {l^{3/4}}\,{n^{1/4}},

{G_2} = {l^{1/2}}{n^{1/2}},\,

\,\,\,\,\,\,\,\,\,\,\,\,\,\,{G_3} = {l^{1/4}}{n^{3/4}}

G_1^4 + 2G_2^4 + G_3^4

= {l^3}n + 2{l^2}{n^2} + {\ln ^3}

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

= \ln \times 2{m^2}

= 4l{m^2}n

Q155

The sum

1 + {{{1^3} + {2^3}} \over {1 + 2}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 2 + 3}} + ...... + {{{1^3} + {2^3} + {3^3} + ... + {{15}^3}} \over {1 + 2 + 3 + ... + 15}}

- {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right)

is equal to :

A 620

B 1240

C 1860

D 660

Correct Answer

Option A

Solution

Sum = \sum\limits_{n = 1}^{15} {{{{1^3} + {2^3} + .... + {n^3}} \over {1 + 2 + .... + n}}} - {1 \over 2}{{15 \times 16} \over 2}

= \sum\limits_{n = 1}^{15} {{{n(n + 1)} \over 2}} - 60

= {1 \over 2}\sum\limits_{n = 1}^{15} {{n^2}} + {1 \over 2}\sum\limits_{n = 1}^{15} {n - 60}

= {1 \over 2} \times {{15 \times 16 \times 31} \over 6} + {1 \over 2} \times {{15 \times 16} \over 2} - 60

= 620

Q156

Let z = 1 + ai be a complex number, a > 0, such that z3 is a real number. Then the sum 1 + z + z2 + . . . . .+ z11 is equal to :

A

- 1250\,\sqrt 3 \,i

B

1250\,\sqrt 3 \,i

C

1365\,\sqrt 3 i

D

-

1365\,\sqrt 3 i

Correct Answer

Option D

Solution

z = 1 + ai z2 = 1 $-$ a2 + 2ai z2 . z = {(1 $-$ a2) + 2ai} {1 + ai} = (1 $-$ a2) + 2ai + (1 $-$ a2) ai $-$ 2a2 $\because$ z3 is real $\Rightarrow$ 2a + (1 $-$ a2) a = 0 a (3 $-$ a2) = 0 $\Rightarrow$ a =

\sqrt 3

(a > 0) 1 + z + z2 . . . . . . . z11 =

{{{z^{12}} - 1} \over {z - 1}} = {{{{\left( {1 + \sqrt 3 i} \right)}^{12}} - 1} \over {1 + \sqrt 3 i - 1}}

=

{{{{\left( {1 + \sqrt 3 i} \right)}^{12}} - 1} \over {\sqrt 3 i}}

(1 +

{\sqrt 3 i}

)12 = 212

{\left( {{1 \over 2} + {{\sqrt 3 } \over 2}i} \right)^{12}}

= 212 (cos

{\pi \over 3}

+ isin

{\pi \over 3}

)12 = 212 (cos4 $\pi$ + isin4 $\pi$ ) = 212 $\Rightarrow$

{{{2^{12}} - 1} \over {\sqrt 3 i}} = {{4095} \over {\sqrt 3 i}} = - {{4095} \over 3}\sqrt 3 i = - 1365\sqrt 3 i

Q157

Let Sn =

{1 \over {{1^3}}}

+ {{1 + 2} \over {{1^3} + {2^3}}} + {{1 + 2 + 3} \over {{1^3} + {2^3} + {3^3}}} + ......... + {{1 + 2 + ....... + n} \over {{1^3} + {2^3} + ...... + {n^3}}}.

If 100 Sn = n, then n is equal to :

A 199

B 99

C 200

D 19

Correct Answer

Option A

Solution

nth term, Tn =

{{1 + 2 + .... + n} \over {{1^2} + {2^2} + .... + {n^2}}}

Tn =

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

$\Rightarrow$ Tn =

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

=

2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]

$\therefore$ Sn =

\sum {{T_n}}

=

2\sum\limits_{n = 1}^n {\left[ {{1 \over n} - {1 \over {n + 1}}} \right]}

=

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

=

{{{2n} \over {n + 1}}}

Given that, 100 Sn = n $\Rightarrow$ 100 $\times$

{{{2n} \over {n + 1}}}

= n $\Rightarrow$ n + 1 = 200 $\Rightarrow$ n = 199

Q158

Let a1, a2, . . . . . ., a10 be a G.P. If

{{{a_3}} \over {{a_1}}} = 25,

then

{{{a_9}} \over {{a_5}}}

equals

A 53

B 2(52)

C 4(52)

D 54

Correct Answer

Option D

Solution

a1, a2, . . . . ., a10 are in G.P., Let the common ratio be r

{{{a_3}} \over {{a_1}}} = 25 \Rightarrow {{{a_1}{r^2}} \over {{a_1}}} = 25 \Rightarrow {r^2} = 25

{{{a_9}} \over {{a_5}}} = {{{a_1}{r^8}} \over {{a_1}{r^4}}} = {r^4} = {5^4}

Q159

If x1, x2, . . ., xn and

{1 \over {{h_1}}}

,

{1 \over {{h_2}}}

, . . . ,

{1 \over {{h_n}}}

are two A.P..s such that x3 = h2 = 8 and x8 = h7 = 20, then x5.h10 equals :

A 2560

B 2650

C 3200

D 1600

Correct Answer

Option A

Solution

Assume d1 is the common difference of A.P x1,x2 ..... xn Given x3 = 8 and x8 = 20 $\therefore$ x1 + 2d1 = 8 ..... (i) and x1 + 7d1 = 20 ..... (ii) Solving (i) and (ii) we get x1 =

16 \over {15}

and d1 =

12 \over {5}

Now let

1 \over d_2

is the common difference of A.P

1 \over h_1

,

1 \over h_2

.....

1 \over h_n

Given that, h2 = 8 and h7 = 20 $\therefore$

1 \over h_2

=

1 \over 8

$\Rightarrow$

1 \over h_1

+

1 \over d_2

=

1 \over 8

.... (iii) and

1 \over h_7

=

1 \over 20

$\Rightarrow$

1 \over h_1

+

6 \over d_2

=

1 \over 20

... (iv) Solving (iii) and (iv) we get

1 \over h_1

=

28 \over 200

and

1 \over d_2

=

- {3 \over 200}

So, x5 = x1 + 4d1 =

16 \over 5

+

48 \over 5

=

64 \over 5

and

1 \over h_{10}

=

1 \over h_1

+

9 \over d_2

=

28 \over 200

-

27 \over 200

=

1\over 200

$\therefore$ x5 $\times$ h10 =

{64 \over 5} \times 200

= 2560

Q160

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12 + 2.22 + 32 + 2.42 + 52 + 2.62 ........... If B - 2A = 100

\lambda

, then

\lambda

is equal to

A 496

B 232

C 248

D 464

Correct Answer

Option C

Solution

Note : Sum of square of first n odd terms 12 + 32 + 52 + . . . . .+ n2 =

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

Given, 12 + 2. 22 + 32 + 2.42 + 52 + 2.62 + . . . . . . A = Sum of first 20 terms

\therefore\,\,\,

A = 12 + 2.22 + 32 + 242 + 52 + 2.62 + . . . . . .20 terms Arrange those terms this way, A = [12 + 32 + 52 + . . . . . 10 terms] + [ 2.22 + 2.42 + 2.62 + . . . . 10 terms] A = [ 12 + 32 + 52 + . . . . 10 terms ] + 2.2 [ 12 + 22 + 32 + . . . .10 terms ] A =

{{10 \times \left( {2.10 - 1} \right)\left( {2.10 + 1} \right)} \over 3} + {2.2^2}\left[ {{{10 \times 11 \times 21} \over 6}} \right]

A =

{{10 \times 19 \times 21} \over 3} + 8 \times {{10 \times 11 \times 21} \over 6}

A =70 $\times$ 19 + 70 $\times$ 44 A = 70 $\times$ 63 B = Sum of first 40 terms Arrange those terms this way.

B = [12+ 32 + 52 +. . . . 20 terms ] + [2.22 + 2.42 +. . . . . 20 terms ] B = [12 + 32 + 52 + . . . . 20 terms] + 2.22 [12 + 22 + . . . 20 terms ] B =

{{20 \times 39 \times 41} \over 3} + \,\,8\,\, \times {{20 \times 21 \times 41} \over 6}

B = 260 $\times$ 41 + 560 $\times$ 41 B = 41

\times \,\,\,820

\therefore\,\,\,

B $-$ 2A = 41

\times \,

820 $-$ 2

\times \,

70

\times \,

63 = 24800 Given that B $-$ 2A = 100 $\lambda$

\therefore\,\,\,

100 $\lambda$ = 24800

\Rightarrow \,\,\,\lambda

= 248