in an arithmetic progression of 15 terms the sum of first 10 terms is 210 and the sum of last 15 term 2565 find the automatic progression

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Assuming your question as :

In an arithmetic progression of 50 terms, the sum of first 10 terms is 210 and the sum of last 15 term 2565 find the arithmetic progression.

Here is the answer.

Let a be the first term and d be the common difference of the given A.P 
We know that sum of first n terms of an A.P is given by,
Sn = n22a + n - 1 dPut n = 10, we get S10 = 1022a + 9d 210 = 5 2a + 9d 2a + 9d = 42   ...........(1)Sum of last 15 terms = 2565Sum of first 50 terms - sum of first 35 terms = 2565S50 - S35 = 25655022a + (50 - 1)d - 3522a + (35 - 1) d = 256525(2a + 49d) - 3522a + 34d = 25655(2a + 49d) - 722a + 34d = 5133a + 126d = 513a + 42d = 171      ...........(2)multiply (2) with '2' we get2a + 84d = 342         ..........(3)subtracting (1) from (3), we get84d - 9d = 342 - 4275d = 300d = 30075 = 4now, from (2), we geta +42(4) =171a + 168=171a = 3

Now, a50 = a + 49d = 3 + 49 × 4 = 3 + 196 = 199
so, the AP is a, a+d, a+2d, a+3d, ........., a+49d

so the required AP is, 3, 7, 11, 15, ......., 199

Regards 

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