The sum of the first $n$ terms of an $A P$ whose first term is $8$ and the common difference is $20$ is equal to the sum of first $2 n$ terms of another $A P$ whose first term is $40$ and the common difference is $6$ . Find $n$ .

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Solution

Let $S_{n}$ be the sum of n terms of an AP with first term a=8 and common difference d=20. Then,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) \Rightarrow S_{n} = \frac{n}{2} (2 \times 8 + (n - 1) \times 20) \Rightarrow S_{n} = \frac{n}{2} (16 + 20 n - 20)$
$\Rightarrow S_{n} = \frac{n}{2} (20 n - 4)$ ...1
similarly Let $S_{2 n}$ be the sum of 2n terms of an AP with first term a1=30 and common difference d1=8. Then,
$S_{2 n} = \frac{2 n}{2} (2 a 1 + (2 n - 1) d 1) \Rightarrow S_{2 n} = \frac{2 n}{2} (2 \times 30 + (n - 1) \times 8) \Rightarrow S_{2 n} = \frac{2 n}{2} (60 + 8 n - 8)$
$\Rightarrow S_{2 n} = n (8 n + 52)$ ...2
According to question $S_{n} = S_{2 n}$
$\frac{n}{2} (20 n - 4) = n (8 n + 52)$
$\Rightarrow 10 n - 2 = 8 n + 52$
$\Rightarrow 2 n = 54$
$\Rightarrow n = 27$

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The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is 40 and the common difference is 6. Find n.

The sum of the first $n$ terms of an $A P$ whose first term is $8$ and the common difference is $20$ is equal to the sum of first $2 n$ terms of another $A P$ whose first term is $40$ and the common difference is $6$ . Find $n$ .