H.P. with (a_1=-\dfrac{1}{3}), (a_5=\dfrac{1}{5})
Reciprocals A.P.: (A_1=-3), (A_5=5)
[ -3+4d=5\implies 4d=8\implies d=2 ]
9th reciprocal: (-3+8\cdot2=13)
9th term of H.P.: (\dfrac{1}{13})
Answer: (\dfrac{1}{13})
11TH•Mathematics•Sequences And Series•Question 3
Overview
The first term of an H.P. is $-\dfrac{1}{3}$ and the fifth term is $\dfrac{1}{5}$. Find its 9th term.
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