The $7^{\text {th }}$ term of an A.P. is 32 and its $13^{\text {th }}$ term is 62 . Find the A.P.
Here, let us take the first term of the A.P. as a and the common difference of the A.P as d
Now, as we know,
$a_{n}=a+(n-1) d$
So, for $7^{\text {th }}$ term $(n=7)$,
$a_{7}=a+(7-1) d$
$32=a+6 d$.........(1)
Also, for 13th term (n = 13),
$a_{13}=a+(13-1) d$
$62=a+12 d$.........(2)
Now, on subtracting (2) from (1), we get,
$62-32=(a+12 d)-(a+6 d)$
$30=a+12 d-a-6 d$
$30=6 d$
$d=\frac{30}{6}$
$d=5$
Substituting the value of d in (1), we get,
$32=a+6(5)$
$32=a+30$
$a=32-30$
$a=2$
So, the first term is 2 and the common difference is 5.
Therefore, the A.P. is $2,7,12,27, \ldots$
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