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(D) The sum of an infinite geometric series is given by $S = \frac{a}{1 - r} = 4$,where $|r| < 1$. The second term of a geometric series is $ar = 3/4$

Vedclass · Accepted Answer

$a = 3, r = 1/4$

Vedclass · Answer

(D) The sum of an infinite geometric series is given by $S = \frac{a}{1 - r} = 4$,where $|r| The second term of a geometric series is $ar = 3/4$. From the second equation,$a = \frac{3}{4r}$. Substituting this into the sum formula: $\frac{3/4r}{1 - r} = 4$. $\frac{3}{4r(1 - r)} = 4$ $3 = 16r(1 - r)$ $3 = 16r - 16r^2$ $16r^2 - 16r + 3 = 0$. Factoring the quadratic equation: $16r^2 - 12r - 4r + 3 = 0$ $4r(4r - 3) - 1(4r - 3) = 0$ $(4r - 1)(4r - 3) = 0$. So,$r = 1/4$ or $r = 3/4$. If $r = 1/4$,then $a = \frac{3}{4(1/4)} = 3$. If $r = 3/4$,then $a = \frac{3}{4(3/4)} = 1$. Checking the options,the pair $(a = 3, r = 1/4)$ matches option $D$.

Consider an infinite geometric series with first term $a$ and common ratio $r$. If its sum is $4$ and the second term is $3/4$,find the values of $a$ and $r$.

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