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 – 30 and the common difference is 8. Find n.

#### Solution

Given that, first term of the first AP (a) = 8

And common difference of the first AP(d) = 20

Let the number of terms in first AP be n

∵ Sum of first n terms of an AP

`S_n = n/2 [2a + (n - 1)d]`

∴ `S_n = n/2[2 xx 8 + (n - 1)20]`

⇒ `S_n = n/2 (16 + 20n - 20)`

⇒ `S_n = n/2(20n - 4)`

∴ `S_n = n(10n - 2)` .....(i)

Now, first term of the second AP(a’) = – 30

And common difference of the second AP(d’) = 8

∴ Sum of first 2n terms of second AP

`S_(2n) = (2n)/2 [2a + (2n - 1)d]`

⇒ S_{2n} = n[2(– 30) + (2n – 1)(8)]

⇒ S_{2n} = n[– 60 + 16n – 8)]

⇒ S_{2n} = n[16n – 68] ………… (ii)

Now, by given condition,

Sum of first n terms of the first AP = Sum of first 2n terms of the second AP

⇒ S_{n} — S_{2n}

⇒ n(10n – 2) = n(16n – 68)

From equation (i) and (ii)

⇒ n[(16n – 68) – (10n – 2)] = 0

⇒ n(16n – 68 – 10n + 2) = 0

⇒ n(6n – 66) = 0

⇒ n = 11 .......[∵ n ≠ 0]

Hence, the required value of n is 11.