The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P.

Question:

The $7^{\text {th }}$ term of an A.P. is 32 and its $13^{\text {th }}$ term is 62 . Find the A.P.

Solution:

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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