Question
Five numbers are in continued proportion. The first term is 5 and the last term is 80 . Find these numbers.
✓
Answer
Let the numbers in continued proportion be $a, a k, a k^2, a k^3, a k^4$.
Here $a=5$ and $a k^4=80$
$
\begin{array}{c}
\therefore 5 \times k^4=80 \\
\therefore k^4=16 \\
\therefore k=2 \quad \because 2^4=16 \\
a k=5 \times 2=10 \quad a k^2=5 \times 4=20 \\
a k^3=5 \times 8=40 \quad a k^4=5 \times 16=80
\end{array}
$
$\therefore$ the numbers are $5,10,20,40,80$.
Here $a=5$ and $a k^4=80$
$
\begin{array}{c}
\therefore 5 \times k^4=80 \\
\therefore k^4=16 \\
\therefore k=2 \quad \because 2^4=16 \\
a k=5 \times 2=10 \quad a k^2=5 \times 4=20 \\
a k^3=5 \times 8=40 \quad a k^4=5 \times 16=80
\end{array}
$
$\therefore$ the numbers are $5,10,20,40,80$.
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