Question

In an ap of 50 terms the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565

Answer 1

Hope this will help you

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 = n2[2a + (n − 1) d]Put n = 10, we get S10 = 102[2a + 9d]⇒ 210 = 5 (2a + 9d)⇒ 2a + 9d = 42   ...........(1)Sum of last 15 terms = 2565⇒Sum of first 50 terms − sum of first 35 terms = 2565⇒S50 − S35 = 2565⇒502[2a + (50 − 1)d] − 352[2a + (35 − 1) d] = 2565⇒25(2a + 49d) − 352(2a + 34d) = 2565⇒5(2a + 49d) − 72(2a + 34d) = 513⇒3a + 126d = 513⇒a + 42d = 171      ...........(2)multiply (2) with '2' we get2a + 84d = 342         ..........(3)subtracting (1) from (3), we get84d − 9d = 342 − 42⇒75d = 300⇒d = 30075 = 4now, from (2), we geta +42(4) =171a + 168=171a = 3

so, the AP is a, a+d, a+2d, a+3d, .........
so the required AP is, 3, 7, 11, 15, .......
 

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Answer 2

$\underline{ \underline{\bf{Answer}}} : - \\ \implies \: 3, \: 7 \:, 11 \: ,15, \: ..........,199 \\ \\ \underline{\underline{ \bf{Step - by - step \: explanation \: }}} : -$

According to the question:-

$\bf{sum \: of \: first \: 10 \: terms \:( s_{10}) = 210} \\ 210 = \frac{10}{2} \bigg (2a + (101)d \bigg) \: \\ \\ 2a + 9d = 42 \: .........(1)\\ \\ \bf{sum \: of \: last \: 15 \: terms \: ( s_{15})= 2565} \\ \\ s_{50} -s_{35} = 2565 \\ \\ 2565 = \frac{50}{2} \bigg(2a + (50 - 1)d \bigg) - \frac{35}{2} \bigg(2a + (35 - 1)d \bigg) \\ \\ 2565 = 25(2a + 49d) - 35(a + 17d) \\ \\ 2565 = 50a + 1225d - 35a - 595d \\ \\ after \: solving \: this \: \\ \\ a + 42d = 171 \: ...........(2) \\ \\ from \: eq(1) \: and \: (2) \\ \\eq (1) \times 42 - \: eq (2) \times 9 \\ \\ we \: get \: \\ \\ a = 3 \: d = 4$

Hence required AP is →

3,7,11,15,....,199