JEE Main 2026 — Jan 24, Shift 1 — Mathematics solution

JEE Main2026Jan 24, Shift 1Mathematics

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 $i = 1 \sum n a_{i} = \frac{525}{2}$ , then $i = 1 \sum 17 a_{i}$ is equal to

A
$476$
B
$952$
C
$238$
D
$136$

Solution

Common difference $d = - 3/4$ . From $a_{n} = a_{1} /4$ :

$a_{1} + (n - 1) d = \frac{a _{1}}{4} \Rightarrow (n - 1) (- \frac{3}{4}) = - \frac{3 a _{1}}{4} \Rightarrow a_{1} = n - 1$ .

Sum to $n$ terms: $\frac{n}{2} (a_{1} + a_{n}) = \frac{n}{2} \cdot \frac{5 a _{1}}{4} = \frac{5 n a _{1}}{8} = \frac{525}{2}$ .

$\Rightarrow n a_{1} = 420$ . Combined with $a_{1} = n - 1$ :

$n (n - 1) = 420 \Rightarrow n^{2} - n - 420 = 0 \Rightarrow n = 21$ , so $a_{1} = 20$ .

Sum to 17 terms: $i = 1 \sum 17 a_{i} = \frac{17}{2} (2 \cdot 20 + 16 \cdot (- \frac{3}{4})) = \frac{17}{2} (40 - 12) = \frac{17}{2} \cdot 28 = 238$ .

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