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1

### 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

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

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

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

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

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

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

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

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

### 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

$$= {1 \over 4}\left[ {{{9 \times 10 \times 19} \over 6} + {{18 \times 10} \over 2} + 9} \right] = {{384} \over 4} = 96$$
3

### JEE Main 2014 (Offline)

If $${(10)^9} + 2{(11)^1}\,({10^8}) + 3{(11)^2}\,{(10)^7} + ......... + 10{(11)^9} = k{(10)^9},$$, then k is equal to :
A
100
B
110
C
$${{121} \over {10}}$$
D
$${{441} \over {100}}$$

## Explanation

Let $${10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ...$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 10{\left( {11} \right)^9} = k{\left( {10} \right)^9}$$

Let $$x = {10^9} + 2.\left( {11} \right){\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ..... + 10{\left( {11} \right)^9}$$

Multiplied by $${{11} \over {10}}$$ on both the sides

$${{11} \over {10}}x = {11.10^8} + 2.{\left( {11} \right)^2}.{\left( {10} \right)^7} + .....$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 9\left( {11} \right){}^9 + {11^{10}}$$

$$x\left( {1 - {{11} \over {10}}} \right) = {10^9} + 11{\left( {10} \right)^8} + 11{}^2 \times {\left( {10} \right)^7}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ... + {11^9} - {11^{10}}$$

$$\Rightarrow - {x \over {10}} = {10^9}\left[ {{{{{\left( {{{11} \over {10}}} \right)}^{10}} - 1} \over {{{11} \over {10}} - 1}}} \right] - {11^{10}}$$

$$\Rightarrow - {x \over {10}} = \left( {{{11}^{10}} - {{10}^{10}}} \right) - {11^{10}} = - {10^{10}}$$

$$\Rightarrow x = {10^{11}} = k{.10^9}$$

Given $$\Rightarrow k = 100$$
4

### JEE Main 2014 (Offline)

Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is :
A
$$2 - \sqrt 3$$
B
$$2 + \sqrt 3$$
C
$$\sqrt 2 + \sqrt 3$$
D
$$3 + \sqrt 2$$

## Explanation

Let $$a,ar,a{r^2}$$ are in $$G.P.$$

According to the question

$$a,2ar,a{r^2}$$ are in $$A.P.$$

$$\Rightarrow 2 \times 2ar = a + a{r^2}$$

$$\Rightarrow 4r = 1 + {r^2}$$

$$\Rightarrow {r^2} - 4r + 1 = 0$$

$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3$$

Since $$r > 1$$

$$\therefore$$ $$\pi = 2 - \sqrt 3$$ is rejected

Hence, $$r = 2 + \sqrt 3$$

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