the sum of first seven terms of an AP is 182, if its 4th term and the 17th term are in ratio 1:5 find the ap
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Given that 4th term and 17th term in ratio 1:5.
(a+3d)/(a+16d) = 1/5
5a + 15d = a + 16d
d = 4a ---- (1).
We know that sum of n terms = n/2(2a+(n-1)*d)
the sum of 7 terms = 7/2(2a + 6d) = 182
= 7(a + 3d) = 182
(a + 3d) = 26.
(a + 3(4a)) = 26 (from (1)
a + 12a = 26
a = 2
Substitute a = 2 in (1)
d = 8.
The Ap = 2,10,18,26,34....
Hope this helps!
(a+3d)/(a+16d) = 1/5
5a + 15d = a + 16d
d = 4a ---- (1).
We know that sum of n terms = n/2(2a+(n-1)*d)
the sum of 7 terms = 7/2(2a + 6d) = 182
= 7(a + 3d) = 182
(a + 3d) = 26.
(a + 3(4a)) = 26 (from (1)
a + 12a = 26
a = 2
Substitute a = 2 in (1)
d = 8.
The Ap = 2,10,18,26,34....
Hope this helps!
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