The numbers lying between 200 and 400 which are divisible by 7

are as follows: -

203, 210, 217, … 399

Since the common difference between the consecutive terms is constant. Thus, the above sequence is an A.P.

∴First term, a = 203

Last term, l = 399

Common difference, d = 7

Let the number of terms of the A.P. be n.

∴ an = 399 = a + (n –1) d

⇒ 399 = 203 + (n –1) 7

⇒ 7 (n –1) = 196

⇒ n –1 = 28

⇒ n = 29

We know that -

Sum of n terms of an A.P(S_n) = (n/2)[a + l]

S₂₉ = (29/2)[203 + 399]

= (29/2)[602]

= 29 × 301

= 8729

Thus, the required sum is 8729.