Question

An ab consist 37th term. the sum of the three middle most term is 225 and last of three 429.find the Ap.

Answer 1

Let the three middle most terms of the AP be: a – d, a, a + d.

So,

(a – d) + a + (a + d) = 225

3a = 225

a = 75

Now, the AP is

a – 18d, …, a – 2d, a – d, a, a + d, a + 2d, …, a + 18d

Sum of last three terms:

(a + 18d) + (a + 17d) + (a + 16d) = 429

3a + 51d = 429

a + 17d = 143

75 + 17d = 143

d = 4

Now, first term = a – 18d = 75 – 18(4) = 3

Therefore, The AP is 3, 7, 11, … , 147.