JEE Main 2018 — Sequences And Series Question with Solution

a₁, a₂, a₃ . . . a₄₃ are in AP

So, a₂ = a₁ + d

a₃ = a₁ + 2d

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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)

$$\sum _{k=0}^{12}{a}_{4k+1}$$ = 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

$$\Rightarrow 13a{}_1+4\times \frac{12\times 13}{2}\times$$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

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a₁₇ = 8 + 16 = 24

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

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

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

We can write above series like this,

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

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

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

$$\Rightarrow$$4760 = 140 m

$$\Rightarrow$$ m = 34