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1

### AIEEE 2009

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

## Explanation

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

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

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

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

$$\Rightarrow S - 3$$
2

### AIEEE 2008

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

## Explanation

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)

$$\Rightarrow a = - 12$$
3

### AIEEE 2007

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

## Explanation

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]
4

### AIEEE 2007

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

## Explanation

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

Put $$x=-1$$

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

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

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