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1

### WB JEE 2010

G.M. and H.M. of two numbers are 10 and 8 respectively. The numbers are

A
5, 20
B
4, 25
C
2, 50
D
1, 100

## Explanation

Let nos. be a and b

(A = A.M., G = G.M., H = H.M.)

Here, G = 10, H = 8

$$\because$$ G2 = AH $$\Rightarrow$$ 100 = 8A

$$\therefore$$ $$A = {{25} \over 2}$$

$$\therefore$$ ab = 100 & a + b = 25, which are satisfied by option (a) only.

2

### WB JEE 2010

Sum of n terms of the following series 13 + 33 + 53 + 73 ......... is

A
n2(2n2 $$-$$ 1)
B
n3(n $$-$$ 1)
C
n3 + 8n + 4
D
2n4 + 3n2

## Explanation

$${1^3} + {3^3} + {5^3} + ....$$ to n terms

$$= {1^3} + {3^3} + {5^3} + .... + {(2n - 1)^3}$$

$$= \{ {1^3} + {2^3} + {3^3} + .... + {(2n)^3}\} - \{ {2^3} + {4^3} + {6^3} + .... + {(2n)^3}\}$$

$$= {\left\{ {{{2n(2n + 1)} \over 2}} \right\}^2} - {2^3}\{ {1^3} + {2^3} + {3^3} + ... + {n^3}\}$$

$$= {n^2}{(2n + 1)^2} - 8{\left\{ {{{n(n + 1)} \over 2}} \right\}^2}$$

$$= {n^2}\{ (4{n^2} + 4n + 1) - 2{(n + 1)^2}\} = {n^2}(2{n^2} - 1)$$

3

### WB JEE 2010

If sum of an infinite geometric series is $${4 \over {3}}$$ and its 1st term is $${3 \over {4}}$$, then its common ratio is

A
$${7 \over {16}}$$
B
$${9 \over {16}}$$
C
$${1 \over 9}$$
D
$${7 \over 9}$$

## Explanation

Here, $$a = {3 \over 4}$$, $${S_\infty } = {4 \over 3}$$

$$\therefore$$ $${a \over {1 - r}} = {4 \over 2} \Rightarrow 3a = 4 - 4r$$

$$\Rightarrow {9 \over 4} = 4 - 4r$$

$$\therefore$$ $$r = {7 \over {16}}$$

4

### WB JEE 2009

The sum of the infinite series $$1 + {1 \over {2!}} + {{1\,.\,3} \over {4!}} + {{1\,.\,3\,.\,5} \over {6!}} + \,....$$ is

A
e
B
e2
C
$$\sqrt e$$
D
1/e

## Explanation

$$1 + {1 \over {2!}} + {{1\,.\,3} \over {4!}} + {{1\,.\,3\,.\,5} \over {6!}} + \,....$$

tn of $${1 \over {2!}} + {{1\,.\,3} \over {4!}} + {{1\,.\,3\,.\,5} \over {6!}} + \,......$$ is

$${t_n} = {{1\,.\,3\,.\,5\,.\,......\,.\,(2n - 1)} \over {(2n)!}}$$

$$= {{1\,.\,2\,.\,3\,.\,4\,.\,5\,.\,6\,.\,......\,(2n - 1)\,.\,(2n)} \over {2\,.\,4\,.\,6\,.\,.....\,(2n)\,(2n)!}}$$

$$= {{(2n)!} \over {(1\,.\,2\,.\,3\,.\,.....\,.\,n)(2n)!}} = {1 \over {{2^n}(n!)}}$$

$$\therefore$$ Sum $$= 1 + \sum\limits_{n = 1}^\infty {{t_n} = 1 + {1 \over {2(1!)}} + {1 \over {{2^2}(2!)}} + {1 \over {{2^3}(3!)}} + .....}$$

$$({e^x} = 1 + {x \over {1!}} + {{{x^2}} \over {2!}} + {{{x^3}} \over {3!}} + .....)$$

$$= {e^{1/2}}$$

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