Question

The sum of all integers
between 100 and 200 which
are divisible by 9 are
O 1677
O 1650
O 1683
O 1680​

Answer 1

Answer:

1683

required sum of the integers between 100 and 200 that are divisible by 9 is 1683.

Answer 2

Answer:

option c

Step-by-step explanation:

first number (between 100 and 200) divisible by 9 is 108

last number divisible by 9 is 198

AP formed = 108,117,126.........198

a = 108

d = 9

Tn = 198

using formula Tn = a+(n-1)d

198 = 108+(n-1)×9

198-108 = 9n-9

90+9 = 9n

99/9 = n

n = 11

now using formula of sum of n terms

Sn = n/2{2a+(n-1)d}

Sn = 11/2{2×108+(11-1)×9}

Sn = 11/2{216+90}

Sn = 11/2 × 306

Sn = 11×153

Sn = 1683

Hence sum will be 1683