JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(a, a r, a r^2, \ldots . . .\). be an infinite \(G.P.\) If \(\sum_{n=0}^{\infty} a^n=57\) and \(\sum_{n=0}^{\infty} a^3 r^{3 n}=9747\), then \(a+18 r\) is equal to :
- A \(27\)
- B \(46\)
- C \(38\)
- D \(31\)
Answer & Solution
Correct Answer
(D) \(31\)
Step-by-step Solution
Detailed explanation
\( \frac{a}{1-r}=57 \) \( \frac{a^3}{1-r^3}=9747 \) \( \frac{a^3}{(1-r)(1+r+r^2)}=9747 \) \( \left(\frac{a}{1-r}\right) \frac{a^2}{1+r+r^2}=9747 \) \( 57 \frac{a^2}{1+r+r^2}=9747 \) \( \frac{a^2}{1+r+r^2}=171 \) \( a=57(1-r) \) \( \frac{(57(1-r))^2}{1+r+r^2}=171 \)…
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