The 3rd and 7th term of a Geometric Progression (GP) are 81 and 16. Find the 5th term.
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Correct Answer: Option D
Explanation:
The nth term of a GP is given by: \(T_{n} = ar^{n-1}\).
\(T_{3} = ar^{3-1} = ar^{2} = 81\).......(1)
\(T_{7} = ar^{7-1} = ar^{6} = 16 \) ...... (2)
Dividing (2) by (1), we have \(r^{4} = \frac{16}{81} = (\frac{2}{3})^{4} \implies r = \frac{2}{3}\)
Putting \(r = \frac{2}{3}\) in equation (1), we have \(81 = a \times (\frac{2}{3}\)^{2} = a \times \frac{4}{9} \implies a = \frac{729}{4}\)
\(T_{5} = ar^{5-1} = ar^{4} = \frac{729}{4} \times (\frac{2}{3})^{4}\)
= \(\frac{729}{4} \times \frac{16}{81} = 36\)
The nth term of a GP is given by: \(T_{n} = ar^{n-1}\).
\(T_{3} = ar^{3-1} = ar^{2} = 81\).......(1)
\(T_{7} = ar^{7-1} = ar^{6} = 16 \) ...... (2)
Dividing (2) by (1), we have \(r^{4} = \frac{16}{81} = (\frac{2}{3})^{4} \implies r = \frac{2}{3}\)
Putting \(r = \frac{2}{3}\) in equation (1), we have \(81 = a \times (\frac{2}{3}\)^{2} = a \times \frac{4}{9} \implies a = \frac{729}{4}\)
\(T_{5} = ar^{5-1} = ar^{4} = \frac{729}{4} \times (\frac{2}{3})^{4}\)
= \(\frac{729}{4} \times \frac{16}{81} = 36\)