JEE Main 2025 Apr 7 Shift 2, Mathematics Q10: Arithmetic, Geometric And Harmonic Progressions

JEE Main2025Apr 7, Shift 2Mathematics

Let $a_{n}$ be the $n^{t h}$ term of an A. P.

If $S_{n} = a_{1} + a_{2} + a_{3} + \dots + a_{n} = 700$ , $a_{6} = 7$ and $S_{7} = 7$ , then $a_{n}$ is equal to :

A
56
B
65
C
64
D
70

Solution

Let the first term be $a$ and common difference $d$ . From $a_{6} = 7$ : $a + 5 d = 7$ . From $S_{7} = 7$ : $\frac{7}{2} (2 a + 6 d) = 7 \Rightarrow a + 3 d = 1$ .

Subtracting, $2 d = 6 \Rightarrow d = 3$ , and then $a = 1 - 9 = - 8$ .

Now use $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = 700$ with $a = - 8, d = 3$ : $\frac{n}{2} [3 n - 19] = 700 \Rightarrow 3 n^{2} - 19 n - 1400 = 0$ .

Solving, $n = \frac{19 \pm 361 + 16800}{6} = \frac{19 \pm 131}{6}$ , giving $n = 25$ (rejecting the negative root).

$∴ a_{25} = a + 24 d = - 8 + 24 (3) = - 8 + 72 = 64$ .

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JEE Main 2025 Apr 7 Shift 2, Mathematics Q10: Arithmetic, Geometric And Harmonic Progressions

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