Question

If a, b, c are the first three terms of a geometric series. If the harmonic mean of a and b is 12 and that of B and C is 36, find the first 5 terms of the series.

Answer 1

Let the geometric series be $a,ar,ar^{2}....$

Given If a, b, c are the first three terms of a geometric series.

So a=a,b=ar and c= $ar^{2}.$

Also harmonic mean of a and b is 12 and that of B and C is 36.

Or $12=\frac{2ab}{a+b}$$12=\frac{2(a)(ar)}{a+ar}=\frac{2a^{2}r }{a(a+r)}=\frac{2ar}{a+r.}$

Or $\frac{ar}{1+r} =6$.........1

Also $\frac{2bc}{b+c}=36.$

$\frac{2ar.ar^{2} }{ar+ar^{2} }$

$\frac{ar^{2} }{a+r}=18.$ .............2

Dividing the two we have:

$\frac{ar}{a+r}.\frac{1+r}{ar^{2} }=\frac{6}{18.}$

Or r=3.

Substituting r value in equation 1

$\frac{a(3)}{1+3}=6$

3a=24

a=8.

The first five terms are :

$a,ar,ar^{2},ar^{3},ar^{4}$

Or 8,24,72,216,648