Question

An AP has 21 terms.The sums of 10,11,12 terms is129.The sum of the last 3 terms is 237.Find the A.P
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Answer 1

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Answer 2

Answer:-

Given:

Number of terms of an AP = 21.

Sum of 10, 11 , 12 terms = 129

We know that,

nth term of an AP = a + (n - 1)d.

→ a(10) + a(11) + a(12) = 129

→ a + (10 - 1)d + a + (11 - 1)d + a + (12 - 1)d = 129

→ 3a + 9d + 10d + 11d = 129

→ 3a + 30d = 129

→ 3(a + 10d) = 129

→ a + 10d = 129/3

→ a + 10d = 43

→ a = 43 - 10d -- equation (1).

And,

Sum of last three terms [ 19 , 20 , 21 terms ] = 237.

→ a + 18d + a + 19d + a + 20d = 237

Substitute the value of "a" here.

→ 43 - 10d + 43 - 10d + 43 - 10d + 57d = 237

→ 129 + 27d = 237

→ 27d = 237 - 129

→ 27d = 108

→ d = 108/27

→ d = 4

Substitute "d" value in equation (1).

→ a = 43 - 10d

→ a = 43 - 10(4)

→ a = 43 - 40

→ a = 3

We know that,

General form of an AP is a , a + d , a + 2d...

Hence , the required AP is 7 , 11 , 15....83.