Question

Find the sum of all numbers from 50 to 250 which divisible by 6 and find t13.

Answer 1

(sum of numbers divisible by 6 from 1 to 250 ) -(sum of numbers divisible from 1 to 50)

=5166-216
=4950

Answer 2

Answer:

13th term is 126 and sum is 4950

Step-by-step explanation:

we have to find the sum of all numbers from 50 to 250 which divisible by 6

The numbers are

$54, 60,66,...246$

The above series forms an A.P with d=6, a=54

First we find the value of n

By recursive formula

$a_n=a+(n-1)d$

$246=54+(n-1)6$

$192=(n-1)6$

$n-1=32$

$n=32+1=33$

we have to find the sum

$S_n=\frac{n}{2}[2a+(n-1)d]$

$S_n=\frac{33}{2}[2(54)+(33-1)6]$

$S_n=\frac{33}{2}[108+192]=4950$

Now, we have to find the 13th term.

$a_{13}=a+12d=54+12(6)=126$

Hence, 13th term is 126 and sum is 4950