QJEEM18Ma90e1: JEE Main 2018 Mathematics - Class 11 JEEM18Ma90e1 Solution: Let , , , ......... , be in A.P. such that and . , then m is…

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