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JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Sequences and Series

Let

S_{k} = \frac{1 + 2 + 3 + \dots .. + k}{k}

. If

S_{1}^{2} + S_{2}^{2} + \dots .. + S_{10}^{2} = \frac{5}{12} A

, then A equal to:

(A)283
(B)301
(C)303✓ Correct
(D)156

Answer: (C)

Solution

First simplify

S_{k} = \frac{1 + 2 + \dots + k}{k} = \frac{k ( k + 1 ) /2}{k} = \frac{k + 1}{2}

. Then

\sum_{k = 1}^{10} S_{k}^{2} = \frac{1}{4} \sum_{k = 1}^{10} (k + 1)^{2} = \frac{1}{4} \sum_{j = 2}^{11} j^{2}

. Using

\sum_{j = 1}^{11} j^{2} = \frac{11 \cdot 12 \cdot 23}{6} = 506

and dropping the

j = 1

term,

\sum_{j = 2}^{11} j^{2} = 505

, so the sum is

\frac{505}{4}

. Setting this equal to

\frac{5}{12} A

gives

A = \frac{505}{4} \cdot \frac{12}{5} = 505 \cdot \frac{3}{5} = 303

.

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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