In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?

Given:

In an A.P., the 5th and 12th terms are 30 and 65 respectively.

To do:

We have to find the sum of the first 20 terms.

Solution:

Let the first term be $a$ and the common differnce be $d$.

Fifth term $a_5=a+(5-1)d$

$30=a+4d$

$a=30-4d$......(i)

12th term $a_{12}=a+(12-1)d$

$65=a+11d$

$65=30-4d+11d$      (From (i))

$7d=65-30$

$d=\frac{35}{7}$

$d=5$.....(ii)

This implies,

$a=30-4(5)$

$=30-20$

$=10$

We know that,

Sum of $n$ terms$ S_{n} =\frac{n}{2}(2a+(n-1)d)$

$S_{20}=\frac{20}{2}[2(10)+(20-1)5]$

$=10(20+95)$

$=10(115)$

$=1150$

Hence, the sum of the first 20 terms is $1150$.