Question

Find the fourth term from the end in an A.P. –11, –8, –5,...., 49.

Answer 1

The given sequence is –11, –8, –5,...., 49.

Here,
First term (a) = –11
Common difference (d) = 3
n = 4
Last term (l) = 49

We know that,
$$\begin{gathered} n^{\mathrm{th}}\mathrm{term}\mathrm{from}\mathrm{the}\mathrm{end}=l-\left(n-1\right)d \\ {4}^{\mathrm{th}}\mathrm{term}\mathrm{from}\mathrm{the}\mathrm{end}=l-\left(4-1\right)d \\ =49-3\left(3\right) \\ =49-9 \\ =40 \end{gathered}$$

Hence, the fourth term from the end is 40.