Question

Consider an infinite geometric series with first term, a and common ratio r. If its sum is 4 and the second terms , is $\frac{3}{4}$ , then a = $\_\_\_\_\_\_$ and r = $\_\_\_\_\_\_$.

• A. $3,\frac{3}{4}$
  
• B. $1,\frac{1}{1}$
• C. $1,\frac{3}{4}$
• D. $2,\frac{3}{4}$

Answer 1

Answer: (A), (C)

The correct options are
A $3,\frac{3}{4}$
C $1,\frac{3}{4}$

$aisfirsttermandriscommonratioifinfiniteG.P.Sum=4\frac{a}{1-r}=4=>a=4-4rGiven{a}^r=\frac{3}{4}4r-4r^2=\frac{3}{4}16r^2-16r+3=0r=\frac{16\pm \sqrt{256-12\times 16}}{2\times 16}r=\frac{16\pm 8}{32}r=\frac{1}{4}\left(or\right)\frac{3}{4}a=3\left(or\right)1a=3,r=\frac{1}{4}\left(or\right)a=1,r=\frac{3}{4}$