Sequences and Series — Page 14 | JEE Mathematics

Q131

If 19th term of a non-zero A.P. is zero, then its (49th term) : (29th term) is :

A 2 : 1

B 4 : 1

C 1 : 3

D 3 : 1

Correct Answer

Option D

Solution

a + 18d = 0 . . . . .(1)

{{a + 48d} \over {a + 28d}} = {{ - 18d + 48d} \over { - 18d + 28d}} = {3 \over 1}

Q132

The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is

{{27} \over {19}}

.Then the common ratio of this series is :

A

{4 \over 9}

B

{1 \over 3}

C

{2 \over 3}

D

{2 \over 9}

Correct Answer

Option C

Solution

{a \over {1 - r}} = 3\,\,\,\,\,\,\,.....(1)

{{{a^3}} \over {1 - {r^3}}} = {{27} \over {19}} \Rightarrow {{27{{\left( {1 - r} \right)}^3}} \over {1 - {r^3}}} = {{27} \over {19}}

\Rightarrow 6{r^2} - 13r + 6 = 0

\Rightarrow r = {2 \over 3}\,\,

as

\left| r \right| < 1

Q133

If the arithmetic mean of two numbers a and b, a > b > 0, is five times their geometric mean, then

{{a + b} \over {a - b}}

is equal to :

A

{{\sqrt 6 } \over 2}

B

{{3\sqrt 2 } \over 4}

C

{{7\sqrt 3 } \over {12}}

D

{{5\sqrt 6 } \over {12}}

Correct Answer

Option D

Solution

A.T.Q., A.M. = 5G.M.

{{a + b} \over 2} = 5\sqrt {ab}

{{a + b} \over {\sqrt {ab} }}

= 10

$\therefore$

{a \over b} = {{10 + \sqrt {96} } \over {10 - \sqrt {96} }} = {{10 + 4\sqrt 6 } \over {10 - 4\sqrt 6 }}

Use componendo and Dividendo

{{a + b} \over {a - b}} = {{20} \over {8\sqrt 6 }} = {5 \over {2\sqrt 6 }} = {{5\sqrt 6 } \over {12}}

Q134

Let

{1 \over {{x_1}}},{1 \over {{x_2}}},...,{1 \over {{x_n}}}\,\,

(xi

\ne

0 for i = 1, 2, ..., n) be in A.P. such that x1=4 and x21 = 20. If n is the least positive integer for which

{x_n} > 50,

then

\sum\limits_{i = 1}^n {\left( {{1 \over {{x_i}}}} \right)}

is equal to :

A

{1 \over 8}

B 3

C

{{13} \over 8}

D

{{13} \over 4}

Correct Answer

Option D

Solution

$\because$

\,\,\,

{1 \over {{x_1}}},{1 \over {{x_2}}},{1 \over {{x_3}}},.....,{1 \over {{x_n}}}

are in A.P. x1 = 4 and x21 = 20 Let 'd' be the common difference of this A.P.

\therefore\,\,\,

its 21st term =

{1 \over {{x_{21}}}} = {1 \over {{x_1}}} + \left[ {\left( {21 - 1} \right) \times d} \right]

$\Rightarrow$

\,\,\,

d =

{1 \over {20}}

$\times$

\left( {{1 \over {20}} - {1 \over 4}} \right)

$\Rightarrow$ d = $-$

{1 \over {100}}

Also xn > 50(given).

\therefore\,\,\,

{1 \over {{x_n}}} = {1 \over {{x_1}}} + \left[ {\left( {n - 1} \right) \times d} \right]

$\Rightarrow$

\,\,\,

xn =

{{{x_1}} \over {1 + \left( {n - 1} \right) \times d \times {x_1}}}

\therefore\,\,\,

{{{x_1}} \over {1 + \left( {n - 1} \right) \times d \times {x_1}}} > 50

$\Rightarrow$

\,\,\,

{4 \over {1 + \left( {n - 1} \right) \times \left( { - {1 \over {100}}} \right) \times 4}} > 50

$\Rightarrow$

\,\,\,

1 + (n $-$ 1) $\times$ ( $-$

{1 \over {100}}

) $\times$ 4 <

{4 \over {50}}

$\Rightarrow$

\,\,\,

$-$

{1 \over {100}}

(n $-$ 1) < $-$

{{23} \over {100}}

$\Rightarrow$

\,\,\,

n $-$ > 23 $\Rightarrow$ n > 24 Therefore

\,\,\,

n = 25. $\Rightarrow$

\,\,\,

\sum\limits_{i = 1}^{25} {{1 \over {{x_i}}}}

=

{{25} \over 2}\left[ {\left( {2 \times {1 \over 4}} \right) + \left( {25 - 1} \right) \times \left( { - {1 \over {100}}} \right)} \right]

=

{{13} \over 4}

Q135

If a, b, c are in A.P. and a2, b2, c2 are in G.P. such that a < b < c and a + b + c =

{3 \over 4},

then the value of a is :

A

{1 \over 4} - {1 \over {4\sqrt 2 }}

B

{1 \over 4} - {1 \over {3\sqrt 2 }}

C

{1 \over 4} - {1 \over {2\sqrt 2 }}

D

{1 \over 4} - {1 \over {\sqrt 2 }}

Correct Answer

Option C

Solution

$\because$

\,\,\,

a, b, c are in A.P. then a + c = 2b also it is given that, a + b + c =

{{3 \over 4}}

. . . .(1) $\Rightarrow$

\,\,\,

2b + b =

{{3 \over 4}}

$\Rightarrow$

\,\,\,

b =

{{1 \over 4}}

. . . . .(2) Again it is given that, a2, b2, c2 are in G.P. then (b2)2 = a2c2 $\Rightarrow$ ac = $\pm$

{{1 \over {16}}}

. . . . (3) From (1), (2) and (3), we get;

a \pm {1 \over {16a}}

=

{1 \over 2}

$\Rightarrow$ 16a2 $-$ 8a $\pm$ 1 = 0 Case I : 16a2 $-$ 8a + 1 = 0 $\Rightarrow$

\,\,\,

a =

{1 \over 4}

(not possible as a < b) Case II: 16a2 $-$ 8a $-$ 1 = 0 $\Rightarrow$

\,\,\,

a =

{{8 \pm \sqrt {128} } \over {32}}

$\Rightarrow$

\,\,\,

a =

{1 \over 4} \pm {1 \over {2\sqrt 2 }}

$\therefore$

\,\,\,

a =

{1 \over 4} - {1 \over {2\sqrt 2 }}

( $\because$ a < b)

Q136

Let An =

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

-

. . . . . + (

-

1)n-1

{\left( {{3 \over 4}} \right)^n}

and Bn = 1

-

An. Then, the least dd natural numbr p, so that Bn > An , for all n

\ge

p, is :

A 9

B 7

C 11

D 5

Correct Answer

Option B

Solution

An =

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

Which in a G.P. with a =

{{3 \over 4}}

, r =

{{{ - 3} \over 4}}

and number of terms = n

\therefore\,\,\,

An =

{{{3 \over 4}\left( {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right)} \over {1 - \left( {{{ - 3} \over 4}} \right)}} = {{{3 \over 4} \times \left( {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right)} \over {{7 \over 4}}}

$\Rightarrow$

\,\,\,

An =

{{3 \over 7}}

\left[ {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right]

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

. . . . . . . . . .(1) As, Bn = 1 $-$ An For least odd natural number p, such that Bn > An $\Rightarrow$

\,\,\,

1 $-$ An > An $\Rightarrow$ 1 > 2 $\times$ An $\Rightarrow$ An <

{{1 \over 2}}

From eqn. (1), we get

{{3 \over 7}}

$\times$

\left[ {1 - {{\left( {{{ - 3} \over 4}} \right)}^n}} \right] < {1 \over 2}

$\Rightarrow$ 1 $-$

{\left( {{{ - 3} \over 4}} \right)^n} < {7 \over 6}

$\Rightarrow$

\,\,\,

1 $-$

{7 \over 6}

<

{\left( {{{ - 3} \over 4}} \right)^n}

$\Rightarrow$

{{ - 1} \over 6} < {\left( {{{ - 3} \over 4}} \right)^n}

As n is odd, then

{\left( {{{ - 3} \over 4}} \right)^n}

= $-$

{{{{3^n}} \over 4}}

So

{{ - 1} \over 6}

< $-$

{\left( {{3 \over 4}} \right)^n}

$\Rightarrow$

{1 \over 6}

>

{\left( {{3 \over 4}} \right)^n}

log

\left( {{1 \over 6}} \right)

= n log

\left( {{3 \over 4}} \right)

$\Rightarrow$ 6.228 < n Hence, n should be 7.

Q137

If A > 0, B > 0 and A + B =

{\pi \over 6}

, then the minimum value of tanA + tanB is :

A

\sqrt 3 - \sqrt 2

B

2 - \sqrt 3

C

4 - 2\sqrt 3

D

{2 \over {\sqrt 3 }}

Correct Answer

Option C

Solution

Given, A + B =

{\pi \over 6}

$\therefore$ tan(A + B) = tan

\left( {{\pi \over 6}} \right)

=

{1 \over {\sqrt 3 }}

We know, tan(A + B) =

{{\tan A + \tan B} \over {1 - \tan A\tan B}}

$\Rightarrow$

{1 \over {\sqrt 3 }}

=

{y \over {1 - \tan A\tan B}}

where y = tan A + tan B $\Rightarrow$ tanA tanB = 1 $-$

\sqrt 3

y Also AM $\ge$ GM $\Rightarrow$

{{\tan A + \tan B} \over 2} \ge \sqrt {\tan A\tan B}

$\Rightarrow$ y $\ge$ 2

\sqrt {1 - \sqrt 3 y}

$\Rightarrow$ y2 $\ge$ 4 $-$ 4

{\sqrt 3 y}

$\Rightarrow$ y2 + 4

{\sqrt 3 y}

$-$ 4 $\ge$ 0 $\Rightarrow$ y $\le$ $-$ 2

\sqrt 3

$-$ 4 or y $\ge$ $-$ 2

\sqrt 3

+ 4 (y $\le$ $-$ 2

\sqrt 3

$-$ 4 is not possible as tan B > 0)

Q138

The sum of the following series

1 + 6 + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right)} \over 9}

+ {{15\left( {{1^2} + {2^2} + ... + {5^2}} \right)} \over {11}} + .....

up to 15 terms, is :

A 7520

B 7510

C 7830

D 7820

Correct Answer

Option D

Solution

1 + 6 + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right)} \over 9} + {{15\left( {{1^2} + {2^2} + ... + {5^2}} \right)} \over {11}} + .....\,15

= {{3\left( {{1^2}} \right)} \over 3} + {{6\left( {{1^2} + {2^2}} \right)} \over 5} + {{9\left( {{1^2} + {2^2} + {3^2}} \right)} \over 7} + {{12} \over 9}\left( {{1^2} + {2^2} + {3^2} + {4^2}} \right) + ......

{T_r} = {{3r} \over {2r + 1}}\left( {{1^2} + {2^2} + .... + {r^2}} \right)

{T_r} = {{3r} \over {2r + 1}}{{r\left( {r + 1} \right)\left( {2r + 1} \right)} \over 6} = {1 \over 2}{r^2}\left( {r + 1} \right)

Sum of

n

terms

= \sum\limits_{r = 1}^n {{T_r}} = {1 \over 2}\sum\limits_{r = 1}^n {\left( {{r^3} + {r^2}} \right)}

= {1 \over 2}\left[ {{{{n^2}{{\left( {n + 1} \right)}^2}} \over 4} + {{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6}} \right]

Sum upto 15 terms $\Rightarrow$ then put

n

= 15

= {1 \over 2}\left( {{{{{\left( {15 \times 16} \right)}^2}} \over 4} + {{15 \times 16 \times 31} \over 6}} \right) = 7820

Q139

Let

3, a, b, c

be in A.P. and

3, a-1, b+1, c+9

be in G.P. Then, the arithmetic mean of

a, b

and

c

is :

A -4

B -1

C 13

D 11

Correct Answer

Option D

Solution

Since $3, a, b, c$ are in arithmetic progression (A.P.), the common difference can be calculated using the term $a$ (the second term) as follows:

d = a - 3

The nth term of an A.P. is given by the formula:

T_n = a + (n-1)d

So, using this formula, we can express $b$ and $c$ in terms of $a$ and $d$ :

b = a + d

c = a + 2d

Substituting $d = a - 3$ into these expressions:

b = a + (a - 3)

c = a + 2(a - 3)

Therefore:

b = 2a - 3

c = 3a - 6

Now, let's consider that $3, a-1, b+1, c+9$ are in geometric progression (G.P.).

For terms in a G.P., the ratio (common ratio, r) between consecutive terms is constant.

So:

\frac{a - 1}{3} = \frac{b + 1}{a - 1} = \frac{c + 9}{b + 1}

Now, we will establish the relation between the terms using the property of G.P.:

\frac{a - 1}{3} = \frac{b + 1}{a - 1}

(a - 1)^2 = 3(b + 1)

a^2 - 2a + 1 = 3b + 3

Substituting $b = 2a - 3$ , we get:

a^2 - 2a + 1 = 3(2a - 3) + 3

a^2 - 2a + 1 = 6a - 9 + 3

a^2 - 8a + 7 = 0

Solving this quadratic equation:

(a - 7)(a - 1) = 0

Hence, $a = 7$ or $a = 1$ .

However, if $a = 1$ , the terms $3, a-1, b+1, c+9$ cannot form a G.P. as it would involve division by zero.

Therefore, $a = 7$ .

We use this value to find $b$ and $c$ :

b = 2a - 3 = 2(7) - 3 = 14 - 3 = 11

c = 3a - 6 = 3(7) - 6 = 21 - 6 = 15

Now we can find the arithmetic mean ( $A$ ) of $a$ , $b$ , and $c$ :

A = \frac{a + b + c}{3}

A = \frac{7 + 11 + 15}{3}

A = \frac{33}{3}

A = 11

Hence, the arithmetic mean of $a$ , $b$ , and $c$ is $11$ , which corresponds to Option D.

Q140

If the sum of the series

\dfrac{1}{1 \cdot(1+\mathrm{d})}+\dfrac{1}{(1+\mathrm{d})(1+2 \mathrm{~d})}+\ldots+\dfrac{1}{(1+9 \mathrm{~d})(1+10 \mathrm{~d})}

is equal to 5, then

50 \mathrm{~d}

is equal to :

A 5

B 10

C 15

D 20

Correct Answer

Option A

Solution

\frac{1}{1 \cdot(1+d)}+\frac{1}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}=5

Multiply and divide by

d

\begin{aligned} & \frac{1}{d}\left[\frac{d}{1 \times(1+d)}+\frac{d}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}\right]=5 \\ & \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\ldots+\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5 \\ & \frac{1}{d}\left[1-\frac{1}{1+10 d}\right]=5 \\ & \frac{1}{d}\left[\frac{1+10 d-1}{1+10 d}\right]=5 \end{aligned}

\begin{aligned} & \frac{10}{1+10 d}=5 \\ & 1+10 d=2 \\ & d=\frac{1}{10} \\ & 50 d=50 \times \frac{1}{10}=5 \end{aligned}