the sum of first 14 terms of an ap is 1505 and its first term is 10 find its 25th term
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\[We have, a = 10 ; n = 14 ; S14 = 1505
Sn = n2[2a + (n − 1)d]⇒1505 = 142[2(10) + (14 − 1)d]⇒1505 = 7[20 + 13d]⇒1505 = 140 + 91d⇒1505 − 140 = 91d⇒d = 136591 = 15
a25 = a + 24d = 10 + 24(15) = 370\]
Sn = n2[2a + (n − 1)d]⇒1505 = 142[2(10) + (14 − 1)d]⇒1505 = 7[20 + 13d]⇒1505 = 140 + 91d⇒1505 − 140 = 91d⇒d = 136591 = 15
a25 = a + 24d = 10 + 24(15) = 370\]
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