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1

AIEEE 2010

MCQ (Single Correct Answer)
A person is to count 4500 currency notes. Let $${a_n}$$ denote the number of notes he counts in the $${n^{th}}$$ minute. If $${a_1}$$ = $${a_2}$$ = ....= $${a_{10}}$$= 150 and $${a_{10}}$$, $${a_{11}}$$,.... are in an AP with common difference - 2, then the time taken by him to count all notes is
A
34 minutes
B
125 minutes
C
135 minutes
D
24 minutes

Explanation

Till $$10$$th minute number of counted notes $$ = 1500$$

$$3000 = {n \over 2}\left[ {2 \times 148 + \left( {n - 1} \right)\left( { - 2} \right)} \right]$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = n\left[ {148 - n + 1} \right]$$

$$ \Rightarrow $$$${n^2} - 149n + 3000 = 0$$

$$ \Rightarrow n = 125,24$$

But $$n=125$$ is not possible

$$\therefore$$ total time $$ = 24 + 10 = 34$$ minutes.
2

AIEEE 2009

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

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

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

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

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

$$ \Rightarrow S - 3$$
3

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)

$$ \Rightarrow a = - 12$$
4

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]

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