Question

Let \(A_{1}, A_{2}, A_{3}, \ldots \ldots\) be an increasing geometric progression. If \(\frac{1}{A_{4}}+\frac{1}{A_{8}}=\frac{10}{63}\) and \(A_{5} A_{7}=441\), then \(3\left(A_{2}+A_{4}\right)\) is equal to ________.

Hint:

\(A_5A_7=A_6^2\), so \(A_6=21\). Write \(A_4=21/r^2\) and \(A_8=21r^2\) and solve for \(r^2\).

Answer 1

Correct Answer: 28

Let the common ratio be \(r\). In a GP, \(A_5A_7=A_6^2=441\), so \(A_6=21\) (the terms must be positive, since \(\frac1{A_4}+\frac1{A_8}>0\)).
\(A_4=\frac{21}{r^2}\), \(A_8=21r^2\).
\(\frac{1}{A_4}+\frac{1}{A_8}=\frac{r^2}{21}+\frac{1}{21r^2}=\frac{10}{63}\Rightarrow r^2+\frac1{r^2}=\frac{10}{3}\Rightarrow 3r^4-10r^2+3=0\Rightarrow r^2=3\) or \(\frac13\).
An increasing GP of positive terms needs \(r>1\), so \(r^2=3\).
\(A_4=7\), \(A_2=\frac{A_4}{r^2}=\frac73\).
\(3(A_2+A_4)=3\left(\frac73+7\right)=7+21=28\).
Answer: 28