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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2008

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

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)

$$ \Rightarrow a = - 12$$
2

AIEEE 2007

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

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]
3

AIEEE 2007

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

Explanation

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

Put $$x=-1$$

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

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

AIEEE 2006

MCQ (Single Correct Answer)
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 $${{{a_6}} \over {a{}_{21}}},\,\,p = 11,\,q = 41$$

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

Questions Asked from Sequences and Series

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