The 16th term of an ap is 5 times its 3rd term. If its 10th term is 41 then find the sum of its first 15 terms
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Hello !
16th term = a + 15d
3rd term = a + 2d
given
5(a + 2d) = a + 15d
5a + 10d = a + 15d
4a - 5d = 0
4a = 5d
a = 5d/4
Also,
10th term = a + 9d = 41
a = 5d/4
so,
5d/4 + 9d = 41
5d/4 + 36d/4 = 41
5d + 36d/4 = 41
41d/4 = 41
41d = 164
d = 164/41
d = 4
a + 9d = 41
a + 36 =41
a = 5
---------------------------------------------
n = 15
a = 5
d = 4
Sn = n/2(2a + (n-1)d)
= 15/2 (10 + 14 * 4)
= 15/2*66
= 495
16th term = a + 15d
3rd term = a + 2d
given
5(a + 2d) = a + 15d
5a + 10d = a + 15d
4a - 5d = 0
4a = 5d
a = 5d/4
Also,
10th term = a + 9d = 41
a = 5d/4
so,
5d/4 + 9d = 41
5d/4 + 36d/4 = 41
5d + 36d/4 = 41
41d/4 = 41
41d = 164
d = 164/41
d = 4
a + 9d = 41
a + 36 =41
a = 5
---------------------------------------------
n = 15
a = 5
d = 4
Sn = n/2(2a + (n-1)d)
= 15/2 (10 + 14 * 4)
= 15/2*66
= 495
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