in an AP of 50 terms , the sum of first 10 terms is 210 and the sum of last 15 terms is 2565. find AP
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n=50
Sum of 10 terms =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.
Hope it is helpful:-D
Answered by
1
According to the question:-
Hence required AP is →
3,7,11,15,....,199
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