Question

In an AP, the first term is 8, nth term is 33 and the sum of first n terms is 123. Thn, d = ?
(a) 5
(b) −5
(c) 7
(d) 3

Answer 1

(a) 5

​​Here, a = 8, T_n = 33 and S_n = 123
Let d be the common difference of the given AP.
Then, T_n = 33
⇒ a + (n - 1)d = 8 + (n-1)d = 33
⇒ (n - 1)d = 25 ...(i)

The sum of n terms of an AP is given by
$$\begin{gathered} S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]=123 \\ \Rightarrow \frac{n}{2}\left[2\times 8+25\right]=\left(\frac{n}{2}\right)\times 41=123 \\ \Rightarrow n=6 \end{gathered}$$ [Substituting the value of (n - 1)d from (i)]
Putting the value of n in (i), we get:
5d = 25
⇒ d = 5