If 36, p,\(\frac{9}{4}\) and q are consecutive terms of an exponential sequence (G.P), find the sum of p and q.
The correct answer is D. 9 \(\frac{9}{16}\)
GP : 36, P, \(\frac{q}{4}\), q, ... p + q = ?
ⴠ\(\frac{1}{4}\) = q ÷ \(\frac{9}{4}\) ;
\(\frac{9}{4}\) = 4q
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| Recall, | common | ratio, | r | = | Tn Tn-1 | = | T2 T1 | = | T3 T2 | = | T4 T3 |
| ⴠ| P 36 | = | 9 4 | ÷ | p | ; | p\(^2\) | = | 9 4 | x | 36 | ; | p\(^2\) | = | 81 |
| p | = | 9 | â´ | r | = | T2 T1 | = | 9 36 | = | 1 4 |
| Also | r | = | T4 T3 | = | q | ÷ | 9 4 |
| 16q | = | 9 | , | q | = | 9 16 | â´ | p | + | q | = | 9 | + | 9 16 | = | 9 | 9 16 |
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