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JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Sequences and Series

Consider an A.P.:

a_{1}, a_{2}, \dots ., a_{n}; a_{1} > 0

. If

a_{2} - a_{1} = \frac{- 3}{4}, a_{n} = \frac{1}{4} a_{1}

, and

\sum_{i = 1}^{n} a_{i} = \frac{525}{2}

, then

\sum_{i = 1}^{17} a_{i}

is equal to

(A)476
(B)952
(C)238✓ Correct
(D)136

Answer: (C)

Solution

With

d = - \frac{3}{4}

and

a_{n} = \frac{a _{1}}{4}

, the sum

S_{n} = \frac{n}{2} (a_{1} + \frac{a _{1}}{4}) = \frac{5 n a _{1}}{8} = \frac{525}{2}

gives

n a_{1} = 420

. Also

a_{n} = a_{1} + (n - 1) d

gives

\frac{a _{1}}{4} = a_{1} - \frac{3}{4} (n - 1)

, i.e.

a_{1} = n - 1

. Then

n (n - 1) = 420

, so

n = 21

and

a_{1} = 20

. Therefore

i = 1 \sum 17 a_{i} = \frac{17}{2} (2 \cdot 20 + 16 \cdot (- \frac{3}{4})) = \frac{17}{2} (40 - 12) = 238

.

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option

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