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in Arithmetic Progression by (41.6k points)
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An A.P. has 21 terms. The sum of 10th, 11th, 12th terms is 129. The sum of the last 3 terms is 237, then find the A.P.

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(a + 9d) + (a + 10d) + (a + 11d) = 129 

3a + 30d = 129 

a + 10d = 43 …………..(1) 

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

3a + 57d = 237

a + 19d = 79 ………………(2)

‘d’ value substituting in equation (1) 

a + 10(4) = 43 

a = 43 – 40 = 3 

∴ a = 3 

∴ Required A.P. is 3, 7, 11, 15, 19,…

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