Question

If a ap consists 21 terms the sum of three middle term is 129 and last tem is 237 find ap

Answer 1

Given total number of terms in an AP = 21

Given that the sum of three terms in the middle is 129

⇒ T10 + T11 + T12 = 129

⇒a + 9d + a + 10d + a + 11d = 129

⇒ 3a + 30d = 129

⇒ a + 10d = 43 --------------------------------------------------(1)

Given that sum of last three terms is 237.

⇒T19 + T20 + T21 = 237

⇒ a + 18d + a + 19d + a + 20d = 237

⇒ 3a + 57d = 237

⇒ a + 19d = 79 -----------------------------------------------(2)

On solving (1) & (2), we get

⇒ a + 10d = 43

⇒ a + 19d = 79

-9d = -36

d = 4

Putting value of d in eq (1), we get

⇒a + 10d = 43

⇒ a + 10(4) = 43

⇒ a + 40 = 43

⇒ a = 43 - 40

⇒ a = 3.

Therefore, The first term is 3 and Common difference = 4.

The required AP is 3,7,11....

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