Question

the 11" term of the A.
ICBS
38. Find the number of all three digit natural numbers which are divisible by 9.
erm.
39. The 19th term of an A.P is equal​

Answer 1

Explanation:

38 solution:

A.P :108,117,129,141..............999

it is an ap

a:108, d: a2-a1 = 117-108=9

let nth term =999

a+(n-1)d=999

108+(n-1)9=999

108+9n-9=999

9n-9=999-108

9n-9=891

9n=891+9

n=900÷9

n= 100

sn=n/2(2a+(n-1)d)

=100/2(2×108+(100-1)9)

=50(216+(99)9

=50(216+891)

=50(1107)

=55350