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
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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
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