1.

Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find is 25th term.

Answer» Here a = 10 and let d be the common difference. Then,
`S_(14) = 1505 rArr (n)/(2) [2a + (n-1)d] = 1505, "where"n = 14 "and" a = 10`
`rArr (14)/(2) * (20 + 13d) = 1505 rArr (20 + 13d) = (1505)/(7) = 215`
`rArr 13d = 195 rArr d = 15.`
Thus,a = 10 and d = 15.
`therefore T_(25) = (a+24d) = (10+24 xx 15) = 370.`
Hence, the 25th term is 370.


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