Question

Find the sum of all multiples of 9 lying between 300 and 700.

Answer 1

The multiples of 9 lying between 300 and 700 are 306, 315,........., 693.

This is an AP with a =  306, d = 9 and l = 693.

Suppose these are n terms in the AP. Then,

a_n = 693  

⇒ 306 + (n – 1) × 9 = 693   ...[a_n = a + (n – 1)d] 

⇒ 9n + 297 = 693

⇒ 9n = 693 – 297

⇒ 9n = 396

⇒ n = 44

∴ Required sum = `44/2 (306 = 693)`   ...`[S_n = n/2 (a + l)]`

= 22 × 999

= 21978

Hence, the required sum is 21978.