Question

An AP consists of 37 terms. The sum of the three middle most term is 225 and sum of last three term is 429. Fimd the A.P.

Answer 1

Hey there!

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

We have, (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

∴ The AP is 3, 7, 11, …, 147.

HOPE IT HELPS ^_^

Answer 2

3 middle terms would be the 18th, 19th and 20th terms

Let these terms be a+ 17d + a + 18d + a + 19d = 225

=> 3a + 54d = 225

=> a + 18d = 75.......i

Sum of the last 3 terms = 429

a + 36d + a+ 35d + a+ 34d = 429

3a + 105d = 429

a + 35d = 143.........ii

subtracting ii from i

17d= 68

d = 4

substituting in i

a= 3

hence the ap is 3, 7,11..............