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