If the second term of an H.P. is 1/5 and the 6th term is 1/13, what is the 10th term?
Answer: B
The corresponding A.P. has a 2nd term of 5 and a 6th term of 13. So, \(a+d=5\) and \(a+5d=13\). Subtracting gives \(4d=8\), so \(d=2\). Substituting d=2 gives \(a+2=5\), so \(a=3\). The 10th term of the A.P. is \(a_{10} = a+9d = 3+9(2) = 3+18=21\). The 10th term of the H.P. is the reciprocal, 1/21.