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,

\[n^{th}\text {  term from the end} = l - \left( n - 1 \right)d\]

\[ 4^{th} \text { term from the end} = l - \left( 4 - 1 \right)d\]

\[ = 49 - 3\left( 3 \right)\]

\[ = 49 - 9\]

\[ = 40\]

Hence, the fourth term from the end is 40.