JEE Mains · Maths · STD 11 - 8. sequence and series
The product of three consecutive terms of a \(G.P.\) is \(512\). If \(4\) is added to each of the first and the second of these terms, the three terms now form an \(A.P.\) Then the sum of the original three terms of the given \(G.P.\) is
- A \(36\)
- B \(32\)
- C \(24\)
- D \(28\)
Answer & Solution
Correct Answer
(D) \(28\)
Step-by-step Solution
Detailed explanation
Let the number be \(\frac{a}{r},a,ar\) Given \({a^3} = 512 \Rightarrow a = 8\) Now given \(\frac{8}{r} + 4,12,8r\) are in \(A.P.\) \( \Rightarrow 2{r^2} - 5r + 2 = 0\) \( \Rightarrow r = \frac{1}{2}\) Numbers are \(4,8,16,\) or \( 16,8,4,\) Sum of numbers \(=4+8+16=28\)
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