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1

### 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]
2

### 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 $$x=-1$$

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

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

### AIEEE 2006

Let $${a_1}$$, $${a_2}$$, $${a_3}$$.....be terms on A.P. If $${{{a_1} + {a_2} + .....{a_p}} \over {{a_1} + {a_2} + .....{a_q}}} = {{{p^2}} \over {{q^2}}},\,p \ne q,\,then\,{{{a_6}} \over {{a_{21}}}}\,$$ equals
A
$${{41} \over {11}}$$
B
$${7 \over 2}$$
C
$${2 \over 7}$$
D
$${{11} \over {41}}$$

## Explanation

$${{{p \over 2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]} \over {{q \over 2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]}} = {{{p^2}} \over {{q^2}}}$$

$$\Rightarrow {{2{a_1} + \left( {p - 1} \right)d} \over {2{a_1} + \left( {p - 1} \right)d}} = {p \over q}$$

$${{{a_1} + \left( {{{p - 1} \over 2}} \right)d} \over {{a_1} + \left( {{{q - 1} \over 2}} \right)d}} = {p \over q}$$

For $${{{a_6}} \over {a{}_{21}}},\,\,p = 11,\,q = 41$$

$$\Rightarrow {{{a_6}} \over {a{}_{21}}} = {{11} \over {41}}$$
4

### AIEEE 2006

If $${{a_1},{a_2},....{a_n}}$$ are in H.P., then the expression $${{a_1}\,{a_2} + \,{a_2}\,{a_3}\, + .... + {a_{n - 1}}\,{a_n}}$$ is equal to
A
$$n({a_1}\, - {a_n})$$
B
$$(n - 1)({a_1}\, - {a_n})$$
C
$$n{a_1}{a_n}$$
D
$$(n - 1)\,\,{a_1}{a_n}$$

## Explanation

$${1 \over {{a_2}}} - {1 \over {{a_1}}} = {1 \over {{a_3}}} - {1 \over {{a_2}}} = .........$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{a_n}}} - {1 \over {{a_{n - 1}}}} = d$$ (say)

Then $${a_1}{a_2} = {{{a_1} - a{}_2} \over d},\,{a_2}{a_3} = {{{a_2} - {a_3}} \over d},$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......,\,{a_{n - 1}}{a_n} = {{{a_{n - 1}} - {a_n}} \over d}$$

$$\therefore$$ $${a_1}a{}_2 + {a_2}{a_3} + ......... + {a_{n - 1}}{a_n}$$

$$= {{{a_1} - {a_2}} \over d} + {{{a_2} - {a_3}} \over d} + ......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {{{a_{n - 1}} - {a_n}} \over d}$$

$$= {1 \over a}\left[ {{a_1}} \right. - {a_2} + {a_2} - {a_3} + .......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + {a_{n - 1}} - an} \right] = {{{a_1} - {a_n}} \over d}$$

Also, $${1 \over {{a_n}}} = {1 \over {{a_1}}} + \left( {n - 1} \right)d$$

$$\Rightarrow {{{a_1} - {a_n}} \over {{a_1}{a_n}}} = \left( {n - 1} \right)d$$

$$\Rightarrow {{{a_1} - {a_n}} \over d} = \left( {n - 1} \right){a_1}{a_n}$$

Which is the required result.

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