Math, asked by chetan123115, 1 year ago

In an ap of 50 terms the sum of first ten terms is 210 And the sum of last fifteen terms of it is 2565 find the ap​

Answers

Answered by vipulsf
5

Answer:

Step-by-step explanation:

Consider a and d as the first term and the common difference of an A.P. respectively.

n th term of an A.P., an = a + ( n – 1)d

Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d]

Given that the sum of the first 10 terms is 210.

⇒ 10 / 2 [2a + 9d ] = 210

⇒ 5[ 2a + 9 d ] = 210

⇒2a + 9d = 42 ----------- (1)

15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning

⇒ a36 = a + 35d

Sum of the last 15 terms = 15/2 [2a36 + ( 15 – 1)d ] = 2565

⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565

⇒ 15 [ a + 35d + 7d ] = 2565

⇒a + 42d = 171 ----------(2)

From (1) and (2), we have d = 4 and a = 3.

Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199.

Answered by Anonymous
3

   \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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