Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

Question Number 43603 by peter frank last updated on 12/Sep/18

If the sum of first two terms of an infinite  GP is 1 and every term is twice the sum  of all the successive terms, then its  first term is

$$\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{two}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an}\:\mathrm{infinite} \\ $$$$\mathrm{GP}\:\mathrm{is}\:\mathrm{1}\:\mathrm{and}\:\mathrm{every}\:\mathrm{term}\:\mathrm{is}\:\mathrm{twice}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{successive}\:\mathrm{terms},\:\mathrm{then}\:\mathrm{its} \\ $$$$\mathrm{first}\:\mathrm{term}\:\mathrm{is} \\ $$

Answered by $@ty@m last updated on 12/Sep/18

a+ar=1  ⇒a=(1/(1+r)) ..(1)  a=2(ar+ar^2 +ar^3 +.....)  ⇒a=((2ar)/(1−r))  ⇒1−r=2r  ⇒r=(1/3) ..(2)  ∴a=(1/(1+(1/3)))   , from(1)  ⇒a =(3/4) Ans.

$${a}+{ar}=\mathrm{1} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{1}+{r}}\:..\left(\mathrm{1}\right) \\ $$$${a}=\mathrm{2}\left({ar}+{ar}^{\mathrm{2}} +{ar}^{\mathrm{3}} +.....\right) \\ $$$$\Rightarrow{a}=\frac{\mathrm{2}{ar}}{\mathrm{1}−{r}} \\ $$$$\Rightarrow\mathrm{1}−{r}=\mathrm{2}{r} \\ $$$$\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{3}}\:..\left(\mathrm{2}\right) \\ $$$$\therefore{a}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}}\:\:\:,\:{from}\left(\mathrm{1}\right) \\ $$$$\Rightarrow{a}\:=\frac{\mathrm{3}}{\mathrm{4}}\:{Ans}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com