JEE Mains · Maths · STD 11 - 8. sequence and series
If each term of a geometric progression \(a_1, a_2, a_3, \ldots\) with \(a_1=\frac{1}{8}\) and \(a_2 \neq a_1\), is the arithmetic mean of the next two terms and \(S_n=a_1+a_2+\ldots+a_n\), then \(\mathrm{S}_{20}-\mathrm{S}_{18}\) is equal to
- A \(2^{ \mathrm{15}}\)
- B \(-2^{18}\)
- C \(2^{18}\)
- D \(-2^{15}\)
Answer & Solution
Correct Answer
(D) \(-2^{15}\)
Step-by-step Solution
Detailed explanation
Let \(r^{\prime}\) th term of the GP be \(a r^{n-1}\). Given, \( 2 a_r=a_{r+1}+a_{r+2} \) \(2 a r^{n-1}=a r^n+a r^{n-1} \) \( \frac{2}{r}=1+r \) \( r^2+r-2=0\) Hence, we get, \(r=-2(\) as \(r \neq 1)\) So, \(\mathrm{S}_{20}-\mathrm{S}_{18}=\) (Sum upto \(20\) terms) - (Sum upto…
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