Question

An AP has 21 terms the sum of 10th,11th and 12th terms of an AP is 129 and sum of the last 3 terms is 237. find the AP

Answer 1

Solution:-
Given : The sum of 10th, 11th and 12th term is 129.
Middle term = (n+1)/2
= (21+1)/2
= 11
The middle term of AP is its 11th term, which is a11
So, the middle three terms are = (a11 - 1),(a11) and (a11 + 1) = a10, a11 and a12.
sum of the 10t, 11th and 12th terms is 129.
so, a10 + a11 + a12 = 129
(a + 9d) + (a +10d) + (a +11d) = 129
3a + 30d = 129 
3(a + 10d) = 129
a + 10d = 129/3
a + 10d = 43 .....................(1)
Also given that the sum of the last three terms is 237.
The last three terms of the AP will be a19, a20 and a21.
So, sum of the last three terms
= a19 + a20 + a21 = 237
(a + 18d) + (a + 19d) + (a + 20d) = 237
3a + 57d = 237
3(a + 19d) = 237
a + 19d = 237/3
a+19d = 79   ......................(2)
Subtracting (1) from (2), we get

  a + 19d = 79.......(2)
  a + 10d = 43.......(1)
-    -           -
_____________
9d = 36
_____________
d = 36/9 
d = 4
Putting the value of d = 4 in the equation (1), we get
a + 10d = 43
a + 10*4 = 43
a = 43 - 40
a = 3
So, a = 3 and d = 4
Therefore, the AP will be 
3, 3 + (1*4), 3 +  (2*4), 3 + (3*4), 3 + (4*4), 3+ (5*4), 3 + (6*4), 3 + (7*4)......
= 3, 7, 11, 15, 19, 23, 27, 32, 35.............
Answer.

Answer 2

a + 9d + a + 10d + a + 11 d = 129 

so, 3a + 30d = 129, 
so, a + 10d = 43. 

similarly, a+18d+a+19d+a+20d = 237, 
3a + 57d = 237
a+19d = 79. 

solving both, we get d = 4. 
a = 3, 

so AP = 3,7,11,15.....79, 83