Let be the term an A.P. If , then is equal to [2025]

Question

Let $a_{n}$ be the $n^{th}$ term of an A.P. If $S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700, a_{6} = 7 and S_{7} = 7$ , then $a_{n}$ is equal to [2025]

Smart Achievers · Accepted Answer

(4)

We have, $S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700, a_{6} = 7 and S_{7} = 7$

Now, $a_{6} = 7 \Rightarrow a + 5 d = 7$ ... (i)

$S_{7} = 7 \Rightarrow \frac{7}{2} (2 a + 6 d) = 7$

$\Rightarrow a + 3 d = 1$ ... (ii)

On solving (i) and (ii), we get

d = 3 and a = – 8

So, $700 = \frac{n}{2} [- 16 + (n - 1) 3]$ $[∵ S_{n} = \frac{n}{2} [2 a + (n - 1) d]]$

$\Rightarrow 1400 = - 16 n + 3 n^{2} - 3 n$

$\Rightarrow 3 n^{2} - 19 n - 1400 = 0$

$\Rightarrow (3 n + 56) (n - 25) = 0 \Rightarrow n = 25$

$∴ a_{25} = a + 24 d = - 8 + 24 \times 3 = 64$ .

Solution

Q.
Let $a_{n}$ be the $n^{th}$ term of an A.P. If $S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700, a_{6} = 7 and S_{7} = 7$ , then $a_{n}$ is equal to [2025]

1 65

2 70

3 56

4 64

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