If \(36, p, \frac{9}{4}, q\) are consecutive terms of an exponential sequence (G.P.). Find the sum of p and q.
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Correct Answer: Option D
Explanation:
\(T_{n} = ar^{n-1}\) (for an exponential sequence)
\(T_{1} = 36 = a\)
\(T_{2} = ar = 36r = p\)
\(T_{3} = ar^{2} = 36r^{2} = \frac{9}{4}\)
\(T_{4} = ar^{3} = 36r^{3} = q\)
\(36r^{2} = \frac{9}{4} \implies r^{2} = \frac{\frac{9}{4}}{36} = \frac{1}{16}\)
\(r = \sqrt{\frac{1}{16}} = \frac{1}{4}\)
\( p = 36 \times \frac{1}{4} = 9 ; q = \frac{9}{4} \times \frac{1}{4} = \frac{9}{16}\)
\(p + q = 9 + \frac{9}{16} = 9\frac{9}{16}\)
\(T_{n} = ar^{n-1}\) (for an exponential sequence)
\(T_{1} = 36 = a\)
\(T_{2} = ar = 36r = p\)
\(T_{3} = ar^{2} = 36r^{2} = \frac{9}{4}\)
\(T_{4} = ar^{3} = 36r^{3} = q\)
\(36r^{2} = \frac{9}{4} \implies r^{2} = \frac{\frac{9}{4}}{36} = \frac{1}{16}\)
\(r = \sqrt{\frac{1}{16}} = \frac{1}{4}\)
\( p = 36 \times \frac{1}{4} = 9 ; q = \frac{9}{4} \times \frac{1}{4} = \frac{9}{16}\)
\(p + q = 9 + \frac{9}{16} = 9\frac{9}{16}\)