Question

An AP consist of 37 terms, the sum of the three middle most term is 225 and the sum of the last three terms of AP is 429...

Find the AP

Answer 1

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 = 19th term.

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

Given a 18 + a 19 + a 20 = 225

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

⇒ 3(a + 18d) = 225

⇒ a + 18d = 75

⇒ a = 75 – 18d … (1)

According to given information

a 35 + a 36 + a 37 = 429

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

⇒ 3(a + 35d) = 429

⇒ (75 – 18d) + 35d = 143

⇒ 17d = 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 …

hope it's correct and helps to you....