Question

An AP consist of 21 terms . The sum of 3 terms in the middle is 129 and of the last 3 is 237 . Find the AP ?

Answer 1

Heya !!

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

We have,

(a-d) + a + (a+d) = 129

=> 3a = 129

=> a = 129/3

=> a = 43

Now, the AP is

a- 10d,..., a-2d, a-d, a, a+d, a+2d,..., a+10d

Sum of the last three terms :

(a+10d) + (a+9d) + (a+8d) = 237

=> 3a + 27d = 237

=> a + 9d = 79

=> 43 + 9d = 79

=> 9d = 79 – 43

=> 9d = 36

=> d = 36/9

=> d = 4

Now, first term = a - 10d

=> 43 –10(4)

=> 3

Therefore, the AP is 3, 7, 11, ..., 83.

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Hope my ans.'s satisfactory.☺