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Question Number 92820 by prince 5 last updated on 13/May/20

a convergent geometric sequence with first term a is such that the sum of the terms after the n^(th) term is three times the n^(th) term, find the common ratio and show that its sum to infinity is 4a.

$${a}\:{convergent}\:{geometric}\:{sequence}\:{with} \\ $$$${first}\:{term}\:{a}\:{is}\:{such}\:{that}\:{the}\:{sum}\:{of} \\ $$$${the}\:{terms}\:{after}\:{the}\:{n}^{{th}} \:{term}\:{is} \\ $$$${three}\:{times}\:{the}\:{n}^{{th}} \:{term},\:{find}\:{the} \\ $$$${common}\:{ratio}\:{and}\:{show}\:{that}\:{its}\: \\ $$$${sum}\:{to}\:{infinity}\:{is}\:\mathrm{4}{a}. \\ $$

Answered by prakash jain last updated on 12/May/20

For a geometric seres fisrt term a common ratio r n^(th) term=ar^(n−1) sum=(a/(1−r)) (∣r∣<1 for covergence) sum after n^(th) term is same as geometric series with first term as (n+1)th term and common ratio r. ((ar^n )/(1−r))=3ar^(n−1) (r/(1−r))=3⇒r=3−3r⇒r=(3/4) Sum of series=(a/(1−r))=4a

$$\mathrm{For}\:\mathrm{a}\:\mathrm{geometric}\:\mathrm{seres}\: \\ $$$$\mathrm{fisrt}\:\mathrm{term}\:{a} \\ $$$$\mathrm{common}\:\mathrm{ratio}\:{r} \\ $$$${n}^{{th}} \:\mathrm{term}={ar}^{{n}−\mathrm{1}} \\ $$$$\mathrm{sum}=\frac{{a}}{\mathrm{1}−{r}}\:\:\:\left(\mid{r}\mid<\mathrm{1}\:\mathrm{for}\:\mathrm{covergence}\right) \\ $$$$\mathrm{sum}\:\mathrm{after}\:{n}^{{th}} \:\mathrm{term}\:\mathrm{is}\:\mathrm{same}\:\mathrm{as} \\ $$$$\mathrm{geometric}\:\mathrm{series}\:\mathrm{with}\:\mathrm{first}\:\mathrm{term} \\ $$$$\mathrm{as}\:\left(\mathrm{n}+\mathrm{1}\right)\mathrm{th}\:\mathrm{term}\:\mathrm{and}\:\mathrm{common}\:\mathrm{ratio}\:{r}. \\ $$$$\frac{{ar}^{{n}} }{\mathrm{1}−{r}}=\mathrm{3}{ar}^{{n}−\mathrm{1}} \\ $$$$\frac{{r}}{\mathrm{1}−{r}}=\mathrm{3}\Rightarrow{r}=\mathrm{3}−\mathrm{3}{r}\Rightarrow{r}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{series}=\frac{{a}}{\mathrm{1}−{r}}=\mathrm{4}{a} \\ $$

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