Question

an AP consists of 21 terms. the sum of the three terms in the middle is 129 and of the last 3 terms is 237. find the AP.

Answer 1

first,
middle 3 terms will be = 10, 11 and 12 th term = 129 (given)
so equation 1 = a+9d + a+10d + a+11d = 129
or we can write 3a + 30d = 129 (Eqn. 1)

Now equation 2 = last 3 terms = 19, 20 and 21st term = 237
so equation 2 = a+18d + a+19d + a+20d = 237
or we can write 3a + 57d = 237

now solve equation 1 and 2
3a + 57d = 237 --- (2)
3a + 30d = 129 --- (1)

after solving the equation we will get common difference (d) as 4 and putting this in equation (1) or (2) we will get first term (a) as 3.

The series will be 3,7,11,15,19...

I hope this answer helps !