Question

An Ap consist of 21terms. The sum of the three terms in the middle is 129 and of the last three is 237.Find the AP​

Answer 1

Solution==>

Let a and d be the first term and common difference of the given A.P.

∴ a 10, a 11, a 12 are the middle terms of the given A.P.

a 10 + a 11 + a 12 = 129

⇒ (a + 9d) + (a + 10d) + (a + 11d) = 129 [an = a + (n – 1)d]

⇒ 3a + 30d = 129

⇒ a + 10d = 43 … (1)

a 19 + a 20 + a 21 = 237

⇒ (a + 18d) + (a + 19d) + (a + 20d) = 237

⇒ 3a + 57d = 237

⇒ a + 19d = 79 … (2)

Solving (1) and (2), we get

a + 19d – a – 10d = 79 – 43

⇒ 9d = 36

⇒ d = 4

When d = 4, we get

a + 10×4 = 43

⇒ a = 43 – 40 = 3

Thus, the given A.P. is 3, 7, 11, 15…

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