4.5
(100k+ )
1

### JEE Main 2016 (Offline)

If the sum of the first ten terms of the series $${\left( {1{3 \over 5}} \right)^2} + {\left( {2{2 \over 5}} \right)^2} + {\left( {3{1 \over 5}} \right)^2} + {4^2} + {\left( {4{4 \over 5}} \right)^2} + .......is\,{{16} \over 5}m,$$ then m is equal to :
A
100
B
99
C
102
D
101

## Explanation

$${\left( {{8 \over 5}} \right)^2} + {\left( {{{12} \over 5}} \right)^2} + {\left( {{{16} \over 5}} \right)^2}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( {{{20} \over 5}} \right)^2}.... + {\left( {{{44} \over 5}} \right)^2}$$

$$S = {{16} \over {25}}\left( {{2^2} + {3^2} + {4^2} + ...... + {{11}^2}} \right)$$

$$= {{16} \over {25}}\left( {{{11\left( {11 + 1} \right)\left( {22 + 1} \right)} \over 6} - 1} \right)$$

$$= {{16} \over {25}} \times 505 = {{16} \over 5} \times 101$$

$$\Rightarrow {{16} \over 5}m = {{16} \over 5} \times 101$$

$$\Rightarrow m = 101.$$
2

### JEE Main 2016 (Offline)

If the $${2^{nd}},{5^{th}}\,and\,{9^{th}}$$ terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is :
A
1
B
$${7 \over 4}$$
C
$${8 \over 5}$$
D
$${4 \over 3}$$

## Explanation

Let the $$GP$$ be $$a,ar$$ and $$a{r^2}$$ then

$$a=A+d;$$ $$ar=A+4d;$$ $$a{r^2} = A + 8d$$

$$\Rightarrow {{a{r^2} - ar} \over {ar - a}} = {{\left( {A + 8d} \right) - \left( {A + 4d} \right)} \over {\left( {A + 4d} \right) - \left( {A + d} \right)}}$$

$$r = {4 \over 3}$$
3

### JEE Main 2015 (Offline)

If m is the A.M. of two distinct real numbers l and n $$(l,n > 1)$$ and $${G_1},{G_2}$$ and $${G_3}$$ are three geometric means between $$l$$ and n, then $$G_1^4\, + 2G_2^4\, + G_3^4$$ equals:
A
$$4\,lm{n^2}$$
B
$$4\,{l^2}{m^2}{n^2}$$
C
$$4\,{l^2}m\,n$$
D
$$4\,l\,{m^2}n$$

## Explanation

$$m = {{l + n} \over 2}$$ and common ratio of

$$G.P.$$ $$= r = {\left( {{n \over l}} \right)^{{1 \over 4}}}$$

$$\therefore$$ $${G_1} = {l^{3/4}}\,{n^{1/4}},$$ $${G_2} = {l^{1/2}}{n^{1/2}},\,$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{G_3} = {l^{1/4}}{n^{3/4}}$$

$$G_1^4 + 2G_2^4 + G_3^4$$

$$= {l^3}n + 2{l^2}{n^2} + {\ln ^3}$$

$$= \ln {\left( {1 + n} \right)^2}$$

$$= \ln \times 2{m^2}$$

$$= 4l{m^2}n$$
4

### JEE Main 2015 (Offline)

The sum of first 9 terms of the series.

$${{{1^3}} \over 1} + {{{1^3} + {2^3}} \over {1 + 3}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 3 + 5}} + ......$$
A
142
B
192
C
71
D
96

## Explanation

$${n^{th}}$$ term of series

$$= {{\left[ {{{n\left( {n + 1} \right)} \over 2}} \right]} \over {{n^2}}} = {1 \over 4}{\left( {n + 1} \right)^2}$$

Sum of $$n$$ term $$= \sum {{1 \over 4}} {\left( {n + 1} \right)^2}$$

$$= {1 \over 4}\left[ {\sum {n{}^2} + 2\sum n + n} \right]$$

$$= {1 \over 4}\left[ {{{n\left( {n + 1} \right)\left( {2n + 1} \right)} \over 6} + {{2n\left( {n + 1} \right)} \over 2} + n} \right]$$

Sum of $$9$$ terms

$$= {1 \over 4}\left[ {{{9 \times 10 \times 19} \over 6} + {{18 \times 10} \over 2} + 9} \right] = {{384} \over 4} = 96$$

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