l, m, n are the $${p^{th}}$$, $${q^{th}}$$ and $${r^{th}}$$ term of a G.P all positive, $$then\,\left| {\matrix{ {\log \,l} & p & 1 \cr {\log \,m} & q & 1 \cr {\log \,n} & r &

$$l = A{R^{p - 1}}$$ $$ \Rightarrow \log 1 = \log A + \left( {p - 1} \right)\log R$$ $$m = A{R^{q - 1}}$$ $$ \Rightarrow \log m = \log A + \left( {q - 1} \right)\log R$$ $$n = A{R^{r - 1}}$$ $$ \Rightarrow \log n = \log A + \left( {r - 1} \right)\log R$$ Now, $$\left| {\matrix{ {\log l} & p & 1 \cr {\log m} & q & 1 \cr {\log n} & r & 1 \cr } } \right|$$ $$ = \left| {\matrix{ {\log A + \left( {p - 1} \right)\log R} & p & 1 \cr {\log A + \left( {q - 1} \right)\log R

Sequences and Series — Page 10 | JEE Mathematics

Q91

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is :

A

2 - \sqrt 3

B

2 + \sqrt 3

C

\sqrt 2 + \sqrt 3

D

3 + \sqrt 2

Correct Answer

Option B

Solution

Let

a,ar,a{r^2}

are in

G.P.

According to the question

a,2ar,a{r^2}

are in

A.P.

\Rightarrow 2 \times 2ar = a + a{r^2}

\Rightarrow 4r = 1 + {r^2}

\Rightarrow {r^2} - 4r + 1 = 0

r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3

Since

r > 1

$\therefore$

\pi = 2 - \sqrt 3

is rejected Hence,

r = 2 + \sqrt 3

Q92

If the sum of the first ten terms of the series

{\left( {1{3 \over 5}} \right)^2} + {\left( {2{2 \over 5}} \right)^2} + {\left( {3{1 \over 5}} \right)^2} + {4^2} + {\left( {4{4 \over 5}} \right)^2} + .......is\,{{16} \over 5}m,

then m is equal to :

A 100

B 99

C 102

D 101

Correct Answer

Option D

Solution

{\left( {{8 \over 5}} \right)^2} + {\left( {{{12} \over 5}} \right)^2} + {\left( {{{16} \over 5}} \right)^2}

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( {{{20} \over 5}} \right)^2}.... + {\left( {{{44} \over 5}} \right)^2}

S = {{16} \over {25}}\left( {{2^2} + {3^2} + {4^2} + ...... + {{11}^2}} \right)

= {{16} \over {25}}\left( {{{11\left( {11 + 1} \right)\left( {22 + 1} \right)} \over 6} - 1} \right)

= {{16} \over {25}} \times 505 = {{16} \over 5} \times 101

\Rightarrow {{16} \over 5}m = {{16} \over 5} \times 101

\Rightarrow m = 101.

Q93

If

{(10)^9} + 2{(11)^1}\,({10^8}) + 3{(11)^2}\,{(10)^7} + ......... + 10{(11)^9} = k{(10)^9},

, then k is equal to :

A 100

B 110

C

{{121} \over {10}}

D

{{441} \over {100}}

Correct Answer

Option A

Solution

Let

{10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ...

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9}

Let

x = {10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7}

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ..... + 10{\left( {11} \right)^9}

Multiplied by

{{11} \over {10}}

on both the sides

{{11} \over {10}}x = {11.10^8} + 2.{\left( {11} \right)^2}.{\left( {10} \right)^7} + .....

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 9\left( {11} \right){}^9 + {11^{10}}

x\left( {1 - {{11} \over {10}}} \right) = {10^9} + 11{\left( {10} \right)^8} + 11{}^2 \times {\left( {10} \right)^7}

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ... + {11^9} - {11^{10}}

\Rightarrow - {x \over {10}} = {10^9}\left[ {{{{{\left( {{{11} \over {10}}} \right)}^{10}} - 1} \over {{{11} \over {10}} - 1}}} \right] - {11^{10}}

\Rightarrow - {x \over {10}} = \left( {{{11}^{10}} - {{10}^{10}}} \right) - {11^{10}} = - {10^{10}}

\Rightarrow x = {10^{11}} = k{.10^9}

Given

\Rightarrow k = 100

Q94

l, m, n are the

{p^{th}}

,

{q^{th}}

and

{r^{th}}

term of a G.P all positive,

then\,\left| \begin{array}{lll}{\log \,l} & p & 1 \\ {\log \,m} & q & 1 \\ {\log \,n} & r & 1 \end{array} \right|\,equals

A - 1

B 2

C 1

D 0

Correct Answer

Option D

Solution

l = A{R^{p - 1}}

\Rightarrow \log 1 = \log A + \left( {p - 1} \right)\log R

m = A{R^{q - 1}}

\Rightarrow \log m = \log A + \left( {q - 1} \right)\log R

n = A{R^{r - 1}}

\Rightarrow \log n = \log A + \left( {r - 1} \right)\log R

Now,

\left| \begin{array}{lll}{\log l} & p & 1 \\ {\log m} & q & 1 \\ {\log n} & r & 1 \end{array} \right|

= \left| \begin{array}{lll}{\log A + \left( {p - 1} \right)\log R} & p & 1 \\ {\log A + \left( {q - 1} \right)\log R} & q & 1 \\ {\log A + \left( {r - 1} \right)\log R} & r & 1 \end{array} \right|

Operating

{C_1} - \left( {\log R} \right){C_2} + \left( {\log R - \log A} \right){C_3}

= \left| \begin{array}{lll}0 & p & 1 \\ 0 & q & 1 \\ 0 & r & 1 \end{array} \right| = 0

Q95

If three distinct numbers a, b, c are in G.P. and the equations ax2 + 2bx + c = 0 and dx2 + 2ex + ƒ = 0 have a common root, then which one of the following statements is correct?

A

d \over a

,

e \over b

,

f \over c

are in G.P.

B d, e, ƒ are in A.P

C d, e, ƒ are in G.P

D

d \over a

,

e \over b

,

f \over c

are in A.P.

Correct Answer

Option D

Solution

Given, a, b, c are in G.P. $\therefore$ b2 = ac In this equation ax2 + 2bx + c = 0, Discrimant, D = 4b2 - 4ac = 4ac - 4ac = 0 Discrimant = 0 meand roots of the equation are equal.

Let both the roots of the equation = $\alpha$ $\therefore$ 2 $\alpha$ =

- {{2b} \over a}

$\Rightarrow$ $\alpha$ =

- {b \over a}

As both the equations ax2 + 2bx + c = 0 and dx2 + 2ex + ƒ = 0 have a common root, so

- {b \over a}

is also root of the equation dx2 + 2ex + ƒ = 0. $\therefore$

- {b \over a}

satisfy the equation dx2 + 2ex + ƒ = 0. $\therefore$

d{\left( { - {b \over a}} \right)^2} + 2e\left( { - {b \over a}} \right) + f = 0

$\Rightarrow$

d{b^2} - 2aeb + f{a^2} = 0

$\Rightarrow$

dac - 2aeb + f{a^2} = 0

$\Rightarrow$

dc - 2eb + fa = 0

$\Rightarrow$

{{dc} \over {ac}} - {{2eb} \over {ac}} + {{fa} \over {ac}} = 0

$\Rightarrow$

{{dc} \over {ac}} - {{2eb} \over {{b^2}}} + {{fa} \over {ac}} = 0

$\Rightarrow$

{d \over a} - {{2e} \over b} + {f \over c} = 0

$\Rightarrow$

{{2e} \over b} = {d \over a} + {f \over c}

$\therefore$

d \over a

,

e \over b

,

f \over c

are in A.P.

Q96

If sin4

\alpha

+ 4 cos4

\beta

+ 2 = 4

\sqrt 2

sin

\alpha

cos

\beta

;

\alpha

,

\beta

\in

[0,

\pi

], then cos(

\alpha

+

\beta

)

-

cos(

\alpha

-

\beta

) is equal to :

A

- \sqrt 2

B 0

C

-

1

D

\sqrt 2

Correct Answer

Option A

Solution

A.M. $\ge$ G.M.

{{{{\sin }^4}\alpha + 4{{\cos }^4}\beta + 1 + 1} \over 4} \ge {\left( {{{\sin }^4}\alpha .4{{\cos }^4}\beta .1.1} \right)^{{1 \over 4}}}

sin4 $\alpha$ + 4 cos2 $\beta$ + 2 $\ge$ 4

\sqrt 2

sin $\alpha$ cos $\beta$ Given that sin4 $\alpha$ + 4cos4 $\beta$ + 2 = 4

\sqrt 2

sin $\alpha$ cos $\beta$ $\Rightarrow$ A.M.

=G.M. $\Rightarrow$ sin4 $\alpha$ = 1 = 4 cos4 $\beta$ sin $\alpha$ = 1, cos $\beta$ = $\pm$

{1 \over {\sqrt 2 }}

$\Rightarrow$ sin $\beta$ =

{1 \over {\sqrt 2 }}

as $\beta$

\in

[0, $\pi$ ] cos( $\alpha$ + $\beta$ ) $-$ cos ( $\alpha$ $-$ $\beta$ ) = $-$ 2 sin $\alpha$ $\beta$ =

- \sqrt 2

Q97

If the sum and product of the first three term in an A.P. are 33 and 1155, respectively, then a value of its 11th term is :-

A –25

B –36

C 25

D –35

Correct Answer

Option A

Solution

Let the three terms are a - d, a, a + d Given a - d + a + a + d = 33 $\Rightarrow$ 3a = 33 $\Rightarrow$ a = 11 Also given, (a - d)a(a + d) = 1155 $\Rightarrow$ (a2 - d2)a = 1155 $\Rightarrow$ (112 - d2)11 = 1155 $\Rightarrow$ (112 - d2) = 105 $\Rightarrow$ d = $\pm$ 4 When d = 4 and a = 11 then series is 7, 11, 15, ....

$\therefore$ T11 = a + 10d = 7 + 10 $\times$ 4 = 47 When d = -4 and a = 11 then series is 15, 11, 7, ....

$\therefore$ T11 = a + 10d = 15 + 10 $\times$ -4 = -25

Q98

In a geometric progression consisting of positive terms, each term equals the sum of the next two terns. Then the common ratio of its progression is equals

A

{\sqrt 5 }

B

\,{1 \over 2}\left( {\sqrt 5 - 1} \right)

C

{1 \over 2}\left( {1 - \sqrt 5 } \right)

D

{1 \over 2}\sqrt 5

.

Correct Answer

Option B

Solution

Let the series

a,ar,

a{r^2},........

are in geometric progression. given,

a = ar + a{r^2}

\Rightarrow 1 = r + {r^2}

\Rightarrow {r^2} + r - 1 = 0

\Rightarrow r = {{ - 1 \mp \sqrt {1 - 4 \times - 1} } \over 2}

\Rightarrow r = {{ - 1 \pm \sqrt 5 } \over 2}

\Rightarrow r = {{\sqrt 5 - 1} \over 2}

[ As terms of

G.P.

are positive $\therefore$

r

should be positive]

Q99

Let Sn denote the sum of the first n terms of an A.P. If S4 = 16 and S6= – 48, then S10 is equal to :

A - 320

B - 380

C - 460

D - 210

Correct Answer

Option A

Solution

S4 =

{4 \over 2}\left( {2a + 3d} \right) = 16

\Rightarrow 2a + 3d = 8

S4 =

{6 \over 2}\left( {2a + 5d} \right) = -48

\Rightarrow 2a + 5d = -16

$\therefore$ d = -12 and a = 22, Now S10 =

{{10} \over 2}\left( {44 - 108} \right) = - 320

Q100

If the sum of the first 20 terms of the series

{\log _{\left( {{7^{1/2}}} \right)}}x + {\log _{\left( {{7^{1/3}}} \right)}}x + {\log _{\left( {{7^{1/4}}} \right)}}x + ...

is 460, then x is equal to :

A e2

B 71/2

C 72

D 746/21

Correct Answer

Option C

Solution

460 = log7 x·(2 + 3 + 4 + ..... + 20 + 21) $\Rightarrow$ 460 = log7 x.

\left( {{{21 \times 22} \over 2} - 1} \right)

$\Rightarrow$ 460 = 230. log7 x $\Rightarrow$ log7 x = 2 $\Rightarrow$ x = 49