Question

AP then is equal toFind the sum of the numbers in
between 1 and 1000 which
divisible by
9​

Answer 1

Given ,

The AP is 9 , 18 , 27 , ...... , 999

Here ,

• First term (a) = 9
• Common difference (d) = 9
• Last term (l) = 999

We know that , the first nth term of an AP is given by

$\mathtt{\large{ \fbox{ a_{n} = a + (n - 1)d}}}$

Thus ,

999 = 9 + (n - 1)9

990 = (n - 1)9

n - 1 = 110

n = 111

We know that , the sum of first n terms of an AP is given by

$\mathtt{ \large{ \fbox{S_{n} = \frac{n}{2} (a + l)}}}$

Thus ,

S = (111/2) × (9 + 999)

S = 111/2 × 998

S = 111 × 499

S = 55389

Hence , the sum is 55389

Answer 2

Answer:

54936

Step-by-step explanation:

Numbers divisible by 9 between 1 to 1000 are

9,18,27..............999

This forms an AP

An=999 , a=9 , d=9 , Sn=? , n=?

An=999

a+(n-1)d=999

9+(n-1)9=999

9n-9=999-9

9n=990-9

n=981/9

n=109

Sn=(n/2)(an+a)

S109=(109/2)(999+9)

=(109/2)(1008)

=(109)(504)

=54936