Consider an infinite geometric series with first term ... | Sequence & Series (Mathematics)

Consider an infinite geometric series with first term

a

and common\n ratio

r

. If its sum is 4 and the second term is

\frac{3}{4}

,\n then

a = \frac{4}{7},r = \frac{3}{7}

a = 2,r = \frac{3}{8}

a = \frac{3}{2},r = \frac{1}{2}

\ a = 3,r = \frac{1}{4}

Correct Answer:

\ a = 3,r = \frac{1}{4}

(d) : We have

\frac{a}{1-r}=4

and ar

=\frac{3}{4}

\nr=\frac{3}{4 a} ; \frac{a}{1-3 / 4 a}=4 \Rightarrow \frac{4 a^2}{4 a-3}=4 \Rightarrow a^2=4 a-3\n

\nor

a^2-4 a+3=0

\n(a-1)(a-3)=0 \Rightarrow a=1 \text { or } 3\n

\n \nWhen

a=1, r=3 / 4

and when

a=3, r=1 / 4

\n\therefore a=3, r=1 / 4\n

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