Question
The $25^{\text {th }}$ term of an A.P. exceeds its $9^{\text {th }}$ term by 16 . Find its common difference.
✓
Answer
$n^{\text {th }}$ term of an A.P. is given by $t_n=a+(n-1) d$.
$\Rightarrow t_{25}=a+(25-1) d=a+24 d \text { and }$
$t_9=a+(9-1) d=a+8 d$
According to the condition in the question, we get
$t_{25}=t_9+16$
$\Rightarrow a+24 d=a+8 d+16$
$\Rightarrow 16 d=16$
$\Rightarrow d=1$
$\Rightarrow t_{25}=a+(25-1) d=a+24 d \text { and }$
$t_9=a+(9-1) d=a+8 d$
According to the condition in the question, we get
$t_{25}=t_9+16$
$\Rightarrow a+24 d=a+8 d+16$
$\Rightarrow 16 d=16$
$\Rightarrow d=1$
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