Math, asked by keerthi234, 1 year ago

in an ap of 50 term the sum of first + 10 term is 210 s the sum of last 15 term is 2565, find the ap

Answers

Answered by yajat1810
1

10/2 [2a+49d] = 210

2a+49d = 42.............(1)

50/2[2a+49d] - 35/2[2a+34d] = sum of last 15 terms

25[2a+49d] - 35[a+17d] = sum of last 15 terms.

25×42 - 35[a+17d] = 2565

1050 - 35[a+17d] = 2565

35[a+17d] = 1050-2565

a+17d = -1515/35

a+17d = -33/7

please mark it as the brainliest answer


keerthi234: tq so much
Answered by Anonymous
0

   \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

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