An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.

Answer:

Let the first term and the common difference of the A.P are *a* and *d* respectively.

Since the A.P contains 37 terms. So, the middle most term is (37+1)/2 th term = 19^{th} term.

Thus, three middle most terms of this A.P.are 18^{th}, 19^{th} and 20^{th} terms.

Given *a* _{18} + *a* _{19} + *a* _{20} = 225

⇒ (*a* + 17*d*) + (*a *+ 18*d*) + (*a *+ 19*d*) = 225

⇒ 3(*a* + 18*d*) = 225

⇒ *a* + 18*d* = 75

⇒ *a* = 75 – 18*d* … (1)

According to given information

*a* _{35} + *a* _{36} + *a* _{37} = 429

⇒ (*a *+ 34*d*) + (*a* + 35*d*) + (*a* + 36*d*) = 429

⇒ 3(*a* + 35*d*) = 429

⇒ (75 – 18*d*) + 35*d* = 143

⇒ 17*d* = 143 – 75 = 68

⇒ *d* = 4

Substituting the value of *d* in equation (1), it is obtained

*a* = 75 – 18 × 4 = 3

Thus, the A.P. is 3, 7, 11, 15 …