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1

MCQ (Single Correct Answer)

The sum to infinite term of the series $$1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + .....$$ is

A

3

B

4

C

6

D

2

We have

$$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + ........\infty \,\,\,\,\,...\left( 1 \right)$$

Multiplying both sides by $${1 \over 3}$$ we get

$${1 \over 3}S = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + {{10} \over {{3^4}}} + .......\,\,\,\,\,...\left( 2 \right)$$

Subtracting eqn. $$(2)$$ from eqn. $$(1)$$ we get

$${2 \over 3}S = 1 + {1 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + {4 \over {{3^4}}} + .....\infty $$

$$ \Rightarrow {2 \over 3}S = {4 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + {4 \over {{3^4}}} + .....\infty $$

$$ \Rightarrow {2 \over 3}S = {{{4 \over 3}} \over {1 - {1 \over 3}}} = {4 \over 3} \times {3 \over 2}$$

$$ \Rightarrow S - 3$$

$$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + {{14} \over {{3^4}}} + ........\infty \,\,\,\,\,...\left( 1 \right)$$

Multiplying both sides by $${1 \over 3}$$ we get

$${1 \over 3}S = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + {{10} \over {{3^4}}} + .......\,\,\,\,\,...\left( 2 \right)$$

Subtracting eqn. $$(2)$$ from eqn. $$(1)$$ we get

$${2 \over 3}S = 1 + {1 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + {4 \over {{3^4}}} + .....\infty $$

$$ \Rightarrow {2 \over 3}S = {4 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + {4 \over {{3^4}}} + .....\infty $$

$$ \Rightarrow {2 \over 3}S = {{{4 \over 3}} \over {1 - {1 \over 3}}} = {4 \over 3} \times {3 \over 2}$$

$$ \Rightarrow S - 3$$

2

MCQ (Single Correct Answer)

The first two terms of a geometric progression add up to 12. the sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is

A

- 4

B

- 12

C

12

D

4

As per question,

$$\,\,\,\,\,\,\,\,\,\,\,\,a + ar = 12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,a{r^2} + a{r^3} = 48\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

$$ \Rightarrow {{a{r^2}\left( {1 + r} \right)} \over {a\left( {1 + r} \right)}} = {{48} \over {12}}$$

$$ \Rightarrow {r^2} = 4, \Rightarrow r = - 2$$

(As terms are $$=+ve$$ and $$-ve$$ alternately)

$$ \Rightarrow a = - 12$$

$$\,\,\,\,\,\,\,\,\,\,\,\,a + ar = 12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,a{r^2} + a{r^3} = 48\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

$$ \Rightarrow {{a{r^2}\left( {1 + r} \right)} \over {a\left( {1 + r} \right)}} = {{48} \over {12}}$$

$$ \Rightarrow {r^2} = 4, \Rightarrow r = - 2$$

(As terms are $$=+ve$$ and $$-ve$$ alternately)

$$ \Rightarrow a = - 12$$

3

MCQ (Single Correct Answer)

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

$${1 \over 2}\sqrt 5 $$.

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]

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]

4

MCQ (Single Correct Answer)

The sum of series $${1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - .......$$ upto infinity is

A

$${e^{ - {1 \over 2}}}$$

B

$${e^{ + {1 \over 2}}}$$

C

$${e^{ - 2}}$$

D

$${e^{ - 1}}$$

We know that $${e^x} = 1 + x + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + ........\infty $$

Put $$x=-1$$

$$\therefore$$ $${e^{ - 1}} = 1 - 1 + {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}}..........\infty $$

$$\therefore$$ $${e^{ - 1}} = {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - {1 \over {5!}}........\infty $$

Put $$x=-1$$

$$\therefore$$ $${e^{ - 1}} = 1 - 1 + {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}}..........\infty $$

$$\therefore$$ $${e^{ - 1}} = {1 \over {2!}} - {1 \over {3!}} + {1 \over {4!}} - {1 \over {5!}}........\infty $$

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