JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4\) be in a geometric progression. \(2,7,9,5\) are subtracted respectively from \(x_1, x_2, x_3\) \(x_4\) then the resulting numbers are in an arithmetic progression. Then the value of \(\frac{1}{24}\left(x_1 x_2 x_3 x_4\right)\) is :
- A 72
- B 18
- C 36
- D 216
Answer & Solution
Correct Answer
(D) 216
Step-by-step Solution
Detailed explanation
\(\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4 \rightarrow \text { G.P. }\) Let a, ar, \(\mathrm{ar}^2, \mathrm{ar}^3 \rightarrow\) G.P. Now \(a-2, a r-7, a^2-9, \mathrm{ar}^3-5 \rightarrow\) A.P. \(2(\mathrm{ar}-7)=\mathrm{a}-2+\mathrm{ar}^2-9\)....(i)…
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