Question

Find three consecutive terms in a G.P. such that the sum of the 2nd and 3rd term is 60 and the product of all the three is 8000.

Answer 1

Let the three consecutive terms in the G.P. be $\frac{a}{r}$, a and ar.
Their product:
$$\begin{gathered} \frac{a}{r}\times a\times ar=8000 \\ \Rightarrow a^3=(20{)}^3 \\ \Rightarrow a=20 \end{gathered}$$

We know that the sum of the 2^nd and 3^rd terms is 60.
Thus, we have:
a + ar = 60
20 + 20r = 30
20r = 30 – 20
20r = 10
r = $\frac{10}{20}=\frac{1}{2}$
∴ $\frac{a}{r}=\frac{20}{{\displaystyle \frac{1}{2}}}=40$, a = 20 and ar = 20 × $\frac{1}{2}$ = 10
Therefore, the three consecutive terms in the G.P. are 10, 20 and 40.