1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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