1

### JEE Main 2017 (Online) 8th April Morning Slot

If the arithmetic mean of two numbers a and b, a > b > 0, is five times their geometric mean, then ${{a + b} \over {a - b}}$ is equal to :
A
${{\sqrt 6 } \over 2}$
B
${{3\sqrt 2 } \over 4}$
C
${{7\sqrt 3 } \over {12}}$
D
${{5\sqrt 6 } \over {12}}$

## Explanation

A.T.Q.,

A.M. = 5G.M.

${{a + b} \over 2} = 5\sqrt {ab}$

${{a + b} \over {\sqrt {ab} }}$ $= 10$

$\therefore$   ${a \over b} = {{10 + \sqrt {96} } \over {10 - \sqrt {96} }} = {{10 + 4\sqrt 6 } \over {10 - 4\sqrt 6 }}$

Use componendo and Dividendo

${{a + b} \over {a - b}} = {{20} \over {8\sqrt 6 }} = {5 \over {2\sqrt 6 }} = {{5\sqrt 6 } \over {12}}$
2

### JEE Main 2017 (Online) 8th April Morning Slot

If the sum of the first n terms of the series $\,\sqrt 3 + \sqrt {75} + \sqrt {243} + \sqrt {507} + ......$ is $435\sqrt 3 ,$ then n equals :
A
18
B
15
C
13
D
29

## Explanation

Given,

$\sqrt 3$ + $\sqrt {75}$ + $\sqrt {243}$ + $\sqrt {507}$ + . . . . . .+ n terms

= $\sqrt 3$ + $\sqrt {25 \times 3}$ + $\sqrt {81 \times 3}$ + $\sqrt {169 \times 3}$ + . . . . . .+ n terms

= $\sqrt 3$ + 5$\sqrt 3$ + 9$\sqrt 3$ + 13$\sqrt 3$ + . . . . . .+ n terms

= $\sqrt 3$ [ 1 + 5 + 9 + 13 + . . . . .+ n terms]

= $\sqrt 3$ $\left[ {{n \over 2}\left( {2.1 + \left( {n - 1} \right)4} \right)} \right]$

= $\sqrt 3$ $\left[ {{n \over 2}\left( {2 + 4n - 4} \right)} \right]$

= $\sqrt 3$ $\left[ {{n \over 2}\left( {4n - 2} \right)} \right]$

= $\sqrt 3$ [n (2n $-$ 1)]

According to question,

$\sqrt 3$ [n (2n $-$ 1)] = 435$\sqrt 3$

$\Rightarrow $$\,\,\, 2n2 - n = 435 \therefore\,\,\, n = {{1 \pm \sqrt {1 + 4 \times 2 \times 435} } \over 4} = {{1 \pm 59} \over 4} \therefore\,\,\, n = {{1 + 59} \over 4} = 15 or {{1 - 59} \over 4} = - 14.5 \therefore\,\,\, n = 15 (as n can't be -ve) 3 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 9th April Morning Slot If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is : A 2 B 4{^{{1 \over 3}}} C 4{^{{2 \over 3}}} D 4 ## Explanation a, b and c are in AP. \therefore a + c = 2b As, abc = 8 \Rightarrow ac\left( {{{a + c} \over 2}} \right)= 8 \Rightarrow ac(a + c) = 16 = 4 \times 4 \therefore ac = 4 and a + c = 4 Then, b = \left( {{{a + c} \over 2}} \right) = {4 \over 2} = 2 4 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 9th April Morning Slot Let Sn = {1 \over {{1^3}}}$$ + {{1 + 2} \over {{1^3} + {2^3}}} + {{1 + 2 + 3} \over {{1^3} + {2^3} + {3^3}}} + ......... + {{1 + 2 + ....... + n} \over {{1^3} + {2^3} + ...... + {n^3}}}.$

If 100 Sn = n, then n is equal to :
A
199
B
99
C
200
D
19

## Explanation

nth term, Tn = ${{1 + 2 + .... + n} \over {{1^2} + {2^2} + .... + {n^2}}}$

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

$\Rightarrow$ Tn = ${2 \over {n\left( {n + 1} \right)}}$ = $2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]$

$\therefore$ Sn = $\sum {{T_n}}$

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

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

= ${{{2n} \over {n + 1}}}$

Given that,

100 Sn = n

$\Rightarrow$ 100 $\times$ ${{{2n} \over {n + 1}}}$ = n

$\Rightarrow$ n + 1 = 200

$\Rightarrow$ n = 199