Question

4th term from the end of an AP -11, -8, -5, -------------49 is​

Answer 1

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AP -11, -8, -5, -------------49

• a=−11
• d=3

$\sf \: a_n =49$

$\sf \: a_n=a+(n−1)d$

$\sf \implies49=−11+(n−1)×3$

$\sf \implies \frac{60}{3} =n−1$

$\sf \implies \: n−1=20$

$\sf \: n=21st \: term$

ㅤ

• Fourth term from the end=18th term

$\sf \: a_{18} =a+(18−1)d$

$\sf \implies \: −11+17×3$

$\sf \implies \: −11+51$

$\sf \implies \: a_{18} =40$