Question

find the sum of integers between hundred and 200 that are divisible by 9
PLZ find using AP formulae

CORRECT AND DETAILED ANSWER WILL BE MARKED AS BRAINLIEST​

Answer 1

Step-by-step explanation:

The sum of integers between 100 and 200 that are divisible by 6 is 2550.

Step-by-step explanation:

First no. between 100 and 200 that is  divisible by 6 is 102

The last no. between 100 and 200 that is  divisible by 6 is 198

Now the numbers between 100 and 200 that is  divisible by 6:

102,102+6,102+6+6 ,....

So, it forma an AP

a = first term = 102

d = common difference = 6

Formula of nth term =

Sum of n terms =

Substitute n =17

Hence the sum of integers between 100 and 200 that are divisible by 6 is 2550.

Answer 2

Between 100 and 200, the first multiple of 9 is 108 and the last multiple of 9 is 198.

Now, we know, the consecutive multiples of a number x are always in AP with common difference(d) x.

According to the given conditions,

• a₁ = 108

• aₙ = 198

• d = 9

We know, aₙ = a₁ + (n - 1)d

⇒198 = 108 + (n - 1)9

⇒90 = 9n - 9

⇒99 = 9n

⇒n = 99/9

⇒n = 11

We also know :-

$\sf{S_n=\dfrac{n}{2}(a+l) }$

Where, n is the number of terms

• a is the first term

• l is the last term

Now, sum of the 11 terms :-

$\sf{S_1_1=\dfrac{11}{2}(108+198) }\\\\\sf{\implies S_1_1=\dfrac{11}{2}(306) }\\\\\sf{\implies S_1_1=11\times153}\\\\\boxed{\sf{\implies S_1_1=1683}}\:\:\:\:\:\:\:\: \cdots \mathbb{\mathbf{ANSWER}}$