MCQ
Consider an infinite $G.P. $ with first term a and common ratio $r$, its sum is $4$ and the second term is $3/4$, then
  • A
    $a = \frac{7}{4},\,r = \frac{3}{7}$
  • B
    $a = \frac{3}{2},\,r = \frac{1}{2}$
  • C
    $a = 2,\,r = \frac{3}{8}$
  • $a = 3,\,r = \frac{1}{4}$

Answer

Correct option: D.
$a = 3,\,r = \frac{1}{4}$
d
(d) Here $\frac{a}{{1 - r}} = 4\,\,{\rm{and}}\,\,ar = \frac{3}{4}$.

Dividing these,

$r(1 - r) = \frac{3}{{16}}$or $16{r^2} - 16r + 3 = 0$

or $(4r - 3)(4r - 1) = 0$

$r = \frac{1}{4},\frac{3}{4}\,\,{\rm{and }}\ a = 3,\,1$

so $(a,r) = \left( {3,\frac{1}{4}} \right)\,,\,\left( {1,\frac{3}{4}} \right)$.

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