An AP consists of 37 terms. The sum of the three middle most terms is . . .

Question : 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.

Doubt by Ananya
Doubt by Saumya

Solution :

Total Number of Terms in AP (n) = 37 (odd)

Middle Term = [(n+1)/2]^th
=[(37+1)/2]^th
=[38/2]^th
=19^th

∴ Three middle most terms are 18th, 19th and 20th

ATQ

a₁₈+a₁₉+a₂₀ = 225
(a+17d)+(a+18d)+(a+19d) = 225
[∵ a_n=(a+(n-1)d]
3a+54d = 225
3(a+18d) = 225
a+18d = 225/3
a+18d = 75 --------(1)

Also,
a₃₅+a₃₆+a₃₇ = 429
(a+34d)+(a+35d)+(a+36d) = 429
3a+105d=429
3(a+35d)=429
a+35d = 429/3
a+35d = 143 -------- (2)

On solving eq (1) and (2) by elimination method

a + 18d =   75
a + 35d = 143
-  -           -  
--------------------
0 - 17d = - 68
--------------------

-17d = -68
17d = 68
d = 68/17
d = 4

Putting in equation (1)
a+18d = 75
a+18 (4) = 75
a+72=75
a=75-72
a=3

∴ The required AP is 

3, 7, 11, 15, 19 . . .