JEE Mains · Maths · STD 11 - 8. sequence and series
Let \( a_{1}, a_{2}, a_{3},..... \) be a G.P. of increasing positive terms such that \( a_{2} . a_{3} . a_{4}=64 \) and \( a_{1}+a_{3}+a_{5}=\frac{813}{7} \).
Then \( a_{3}+a_{5}+a_{7} \) is equal to:
- A 3256
- B 3252
- C 3244
- D 3248
Answer & Solution
Correct Answer
(B) 3252
Step-by-step Solution
Detailed explanation
\( ar . ar^{2}.ar^{3}=64 \) \( a^{3}r^{6}=64\Rightarrow ar^{2}=4 \) \( a+ar^{2}+ar^{4}=\frac{813}{7} \) \( r^{2}=28 \) \( ar^{2}+ar^{4}+ar^{6}= ?\) \(\operatorname{ar}^2\left(1+r^2+r^4\right)=4(1+28+784)=3252\)
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