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### 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 $$\,\,\, 2n2 - n = 435 \therefore\,\,\, n = {{1 \pm \sqrt {1 + 4 \times 2 \times 435} } \over 4} = {{1 \pm 59} \over 4} \therefore\,\,\, n = {{1 + 59} \over 4} = 15 or {{1 - 59} \over 4} = - 14.5 \therefore\,\,\, n = 15 (as n can't be -ve) 2 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 9th April Morning Slot If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is : A 2 B 4{^{{1 \over 3}}} C 4{^{{2 \over 3}}} D 4 ## Explanation a, b and c are in AP. \therefore a + c = 2b As, abc = 8 \Rightarrow ac\left( {{{a + c} \over 2}} \right)= 8 \Rightarrow ac(a + c) = 16 = 4 \times 4 \therefore ac = 4 and a + c = 4 Then, b = \left( {{{a + c} \over 2}} \right) = {4 \over 2} = 2 3 MCQ (Single Correct Answer) ### JEE Main 2017 (Online) 9th April Morning Slot Let Sn = {1 \over {{1^3}}}$$ + {{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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### JEE Main 2018 (Offline)

Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series
12 + 2.22 + 32 + 2.42 + 52 + 2.62 ...........
If B - 2A = 100$\lambda$, then $\lambda$ is equal to
A
496
B
232
C
248
D
464

## Explanation

Note :

Sum of square of first n odd terms

12 + 32 + 52 + . . . . .+ n2 = ${{n\left( {2n - 1} \right)\left( {2n + 1} \right)} \over 3}$

Given,

12 + 2. 22 + 32 + 2.42 + 52 + 2.62 + . . . . . .

A = Sum of first 20 terms

$\therefore\,\,\,$A = 12 + 2.22 + 32 + 242 + 52 + 2.62 + . . . . . .20 terms

Arrange those terms this way,

A = [12 + 32 + 52 + . . . . . 10 terms] + [ 2.22 + 2.42 + 2.62 + . . . . 10 terms]

A = [ 12 + 32 + 52 + . . . . 10 terms ] + 2.2 [ 12 + 22 + 32 + . . . .10 terms ]

A = ${{10 \times \left( {2.10 - 1} \right)\left( {2.10 + 1} \right)} \over 3} + {2.2^2}\left[ {{{10 \times 11 \times 21} \over 6}} \right]$

A = ${{10 \times 19 \times 21} \over 3} + 8 \times {{10 \times 11 \times 21} \over 6}$

A =70 $\times$ 19 + 70 $\times$ 44

A = 70 $\times$ 63

B = Sum of first 40 terms

Arrange those terms this way.

B = [12+ 32 + 52 +. . . . 20 terms ] + [2.22 + 2.42 +. . . . . 20 terms ]

B = [12 + 32 + 52 + . . . . 20 terms] + 2.22 [12 + 22 + . . . 20 terms ]

B = ${{20 \times 39 \times 41} \over 3} + \,\,8\,\, \times {{20 \times 21 \times 41} \over 6}$

B = 260 $\times$ 41 + 560 $\times$ 41

B = 41 $\times \,\,\,820$

$\therefore\,\,\,$ B $-$ 2A = 41 $\times \,$ 820 $-$ 2 $\times \,$ 70 $\times \,$ 63 = 24800

Given that B $-$ 2A = 100 $\lambda$

$\therefore\,\,\,$ 100 $\lambda$ = 24800

$\Rightarrow \,\,\,\lambda$ = 248

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