1

### 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
2

### 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
3

### 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
4

### JEE Main 2018 (Offline)

Let ${a_1}$, ${a_2}$, ${a_3}$, ......... ,${a_{49}}$ be in A.P. such that

$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$ and ${a_9} + {a_{43}} = 66$.

$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$, then m is equal to
A
33
B
66
C
68
D
34

## Explanation

a1, a2, a3 . . . a43 are in AP

So, a2 = a1 + d

a3 = a1 + 2d

.

.

.

a49 =a1 + 48d

Now given, ${a_9} + {a_{43}} = 66$

$\Rightarrow \,\,\,\,$ a1 + 8d + a1 + 42d = 66

$\Rightarrow \,\,\,\,$ 2a1 + 50d = 66

$\Rightarrow \,\,\,\,$ a1 + 25d = 33 . . . . . (1)

$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}}$ = 416

$\Rightarrow \,\,\,\,$ a1 + a5 + a9 + a13 +. . . . . 13 items = 416

$\Rightarrow \,\,\,\,$ a1 + a1 + 4d + a1 + 8d + . . . . a1 + 48d = 416

$\Rightarrow \,\,\,\,$ 13a1 + 4d +8d + 12d + . . . . . 48d = 416

$\Rightarrow \,\,\,\,$ 13a1 + 4 (1+ 2 + 3 + . . . + 12) d = 416

$\Rightarrow \,\,\,\,13\,\,a{}_1 + \,4\,\, \times \,{{12 \times 13} \over 2} \times$d = 416

$\Rightarrow \,\,\,\,$ 13a1 + 24 $\times$ 13d = 416

$\Rightarrow \,\,\,\,$ a1 + 24 d =32 . . . .(2)

Solving (1) and (2) we get,

d = 1

and ${a_1} = 8$

$\therefore\,\,\,$ a1 = 8

a2 = 8 + 1 = 9

a3 = 8 + 2 = 10

.

.

.

a17 = 8 + 16 = 24

Now, $a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2\,\, = \,\,140m$

$\Rightarrow \,\,\,\,$ $a_1^2 + a{}_2^2 + ......\,\, + a_{17}^2 = 140\,m$

$\Rightarrow \,\,\,\,\,{8^2}\, + \,\,{9^2}\, + \,{10^2} + ......{(24)^2} = 140\,m$

We can write above series like this,

$\Rightarrow \,\,\,\,\,$ (12 +22 + . . . . +242) $-$ (12 + 22 + . . . . .+ 72) = 140 m

$\Rightarrow {{24\left( {25} \right)\left( {49} \right)} \over 6} - {{7 \times 8 \times 15} \over 6} = 140\,m$

$\Rightarrow \,\,\,\,\,$ 490 $-$ 140 = 140 m

$\Rightarrow \,\,\,\,\,$4760 = 140 m

$\Rightarrow \,\,\,\,\,$ m = 34