Question

if the three consecutive terms of a gp is increased by their middle term then prove that the resulting terms will be in Hp

Answer 1

Answer:

Let a be the first term of the GP and r the common ratio.  Then the three terms of the GP are a, ar, ar².

Adding ar (the middle term) to each of these gives:

a+ar, ar+ar, ar²+ar                      ... (1)

or equivalently

a ( 1 + r ),  2ar,  ar ( 1 + r ).             ... (1*)

To show that these are in HP, we need to show that reciprocals

1 / a(1 + r),  1 / 2ar,  1 / ar(1 + r)        ... (2)

are in AP.  This follows once we see that the middle term here is the (arithmetic) mean of the other two terms.

Now...

$\displaystyle\frac12\left(\frac1{a(1+r)} + \frac1{ar(1+r)}\right)\\\\=\frac12\left(\frac{r}{ar(1+r)}+\frac1{ar(1+r)}\right)\\\\=\frac12\times\frac{r+1}{ar(1+r)}\\\\=\frac1{2ar}$

Therefore the terms in sequence (2) are in AP, and so the terms in sequence (1*)  [ or equivalently, in sequence (1) ] are in HP, as required.

Answer 2

Answer:

Step-by-step explanation: