Question

Three numbers are h.p such that the sum of whose reciprocals is 12 and product is 1/48

Answer 1

Answer:

Step-by-step explanation:

Let the H.P. be 1/a + 1/(a + d) + 1(a + 2d) + ...

The 7th term = 1/(a + 6d) = 1/10 =>  a +6 d = 10

The 12th term = 1/(a + 11d) = 1/25  => a +11 d = 25

Solving these two equations, a = -8, d = 3

Hence 20th term = 1/(a+19d) = 1/[-8 + 9(3)] = 1/49

and nth term = 1/[a +(n -1)d] = 1/[-8 +(n -1) 3] = 1/[3n - 11]

Answer 2

Answer:

here is what can help u

Let the H.P. be 1/a + 1/(a + d) + 1(a + 2d) + ...

The 7th term = 1/(a + 6d) = 1/10 => a +6 d = 10

The 12th term = 1/(a + 11d) = 1/25 => a +11 d = 25

Solving these two equations, a = -8, d = 3

Hence 20th term = 1/(a+19d) = 1/[-8 + 9(3)] = 1/49

and nth term = 1/[a +(n -1)d] = 1/[-8 +(n -1) 3] = 1/[3n - 11]

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