Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25th term.
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Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25th term.
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hey,
here your answer....
Sum of n terms = Sn = n/2 [ 2a + (n-1) d ]
where a = first term d = common difference
therefore, S14 = 14/2 [2*10 + (14-1) d ] = 1505
solve for d, we get, 215 = 20 + 13d
hence, d = 15.
now, an = a + (n-1) d
therefore, a25 = 10 + (25-1) 15
a25 = 10 + 24 * 15 = 370.
your answer is 370
hope this helps.
:)
Given First term = 10.
Given sum of first 14 terms = 1505.
We know that sum of first n terms of an AP sn = (n/2)[2a + (n - 1) * d]
⇒ 1505 = (14/2)[20 + 13 * d]
⇒ 3010 = 14[20 + 13d]
⇒ 215 = 20 + 13d
⇒ 195 = 13d
⇒ d = 15.
We know that nth term of an AP an = a + (n - 1) * d
25th term: a + (25 - 1) * d
⇒ 10 + 24 * (15)
⇒ 10 + 360
⇒ 370.
Therefore, its 25th term = 370.
Hope it helps!