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Q. The $30^{th}$ term of the arithmetic progression $10, 7, 4$ is

KEAMKEAM 2017Sequences and Series

Solution:

We have, $10,7,4, \ldots \ldots$
which is an $A.P$.
$\therefore a=10, d=-3$
$\therefore a_{30}=a+29 \,d \,\,\,\left[\because a_{n}=a+(n-1) d\right]$
$=10+29(-3)$
$=10-87$
$=-77$