Q26 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
Consider an A.P.: $a_{1}, a_{2}, \ldots, a_{\mathrm{n}} ; a_{1}>0$. If $a_{2}-a_{1}=\frac{-3}{4}, a_{\mathrm{n}}=\frac{1}{4} a_{1}$, and $\sum_{\mathrm{i}=1}^{\mathrm{n}} a_{\mathrm{i}}=\frac{525}{2}$, then $\sum_{\mathrm{i}=1}^{17} a_{\mathrm{i}}$ is equal to
- A136
- B476
- Correct answerC238
- D952
Solution
Common difference $d = a_2 - a_1 = -3/4$.
$a_n = a_1 + (n-1)(-3/4) = a_1/4$
$\Rightarrow$ $3a_1/4 = 3(n-1)/4$
$\Rightarrow a_1 = n-1$.
$S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}\cdot\frac{5a_1}{4} = \frac{5na_1}{8} = \frac{525}{2}$
$na_1 = 420$.
Since $a_1 = n-1$: $n(n-1) = 420$
$\Rightarrow n = 21,\; a_1 = 20$.
$\sum_{i=1}^{17} a_i = \frac{17}{2}(2 \cdot 20 + 16 \cdot (-3/4))$
$= \frac{17}{2}(40 - 12) = 17 \times 14 = 238$
$a_n = a_1 + (n-1)(-3/4) = a_1/4$
$\Rightarrow$ $3a_1/4 = 3(n-1)/4$
$\Rightarrow a_1 = n-1$.
$S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}\cdot\frac{5a_1}{4} = \frac{5na_1}{8} = \frac{525}{2}$
$na_1 = 420$.
Since $a_1 = n-1$: $n(n-1) = 420$
$\Rightarrow n = 21,\; a_1 = 20$.
$\sum_{i=1}^{17} a_i = \frac{17}{2}(2 \cdot 20 + 16 \cdot (-3/4))$
$= \frac{17}{2}(40 - 12) = 17 \times 14 = 238$
Hence, the correct answer is option: 'C'.