Let a_1, a_2, a_3, ldots be a G.P. of increas | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the $G.P.$ be $a, ar, ar^2, ar^3, ar^4, ar^5, \ldots$ where $a > 0$ and $r > 1$.Given $a_1 a_5 = 28$ $\Rightarrow a(ar^4) = 28$ $\Rightarrow a

Vedclass · Accepted Answer

$784$

Vedclass · Answer

(C) Let the $G.P.$ be $a, ar, ar^2, ar^3, ar^4, ar^5, \ldots$ where $a > 0$ and $r > 1$.Given $a_1 a_5 = 28$ $\Rightarrow a(ar^4) = 28$ $\Rightarrow a^2 r^4 = 28$ $\Rightarrow (ar^2)^2 = 28$ $\Rightarrow ar^2 = \sqrt{28} = 2\sqrt{7}$.Given $a_2 + a_4 = 29$ $\Rightarrow ar + ar^3 = 29$ $\Rightarrow ar(1 + r^2) = 29$.Since $ar^2 = \sqrt{28}$,we have $a = \frac{\sqrt{28}}{r^2}$.Substituting $a$ in the second equation: $\frac{\sqrt{28}}{r^2} \cdot r(1 + r^2) = 29 \Rightarrow \frac{\sqrt{28}(1 + r^2)}{r} = 29$.Let $x = r + \frac{1}{r}$. Then $r^2 + 1 = xr$.The equation becomes $\sqrt{28} \cdot x = 29 \Rightarrow x = \frac{29}{\sqrt{28}}$.Since $r + \frac{1}{r} = \frac{29}{\sqrt{28}}$,we solve for $r$: $r^2 - \frac{29}{\sqrt{28}}r + 1 = 0$.Using the quadratic formula,$r = \frac{\frac{29}{\sqrt{28}} \pm \sqrt{\frac{841}{28} - 4}}{2} = \frac{\frac{29}{\sqrt{28}} \pm \sqrt{\frac{729}{28}}}{2} = \frac{\frac{29 \pm 27}{\sqrt{28}}}{2}$.For increasing terms,$r > 1$,so $r = \frac{56}{2\sqrt{28}} = \frac{28}{\sqrt{28}} = \sqrt{28}$.Then $a = \frac{\sqrt{28}}{r^2} = \frac{\sqrt{28}}{28} = \frac{1}{\sqrt{28}}$.Finally,$a_6 = ar^5 = \frac{1}{\sqrt{28}} \cdot (\sqrt{28})^5 = (\sqrt{28})^4 = 28^2 = 784$.

Let $a_1, a_2, a_3, \ldots$ be a $G.P.$ of increasing positive terms. If $a_1 a_5 = 28$ and $a_2 + a_4 = 29$,then $a_6$ is equal to

Similar Questions

If the third term of a $G.P.$ is $4$,then the product of its first $5$ terms is:

If the roots of the equation $8x^3 - 14x^2 + 7x - 1 = 0$ are in $G.P.$,then the roots are

If $\alpha, \beta, \gamma$ are the roots of the equation $3x^3-26x^2+52x-24=0$ such that $\alpha, \beta, \gamma$ are in geometric progression and $\alpha < \beta < \gamma$,then $3\alpha + 2\beta + \gamma =$

The $7^{th}$ term of the sequence $\sqrt{2}, \sqrt{10}, 5\sqrt{2}, \dots$ is

If $S_n$ stands for the sum of $n$ terms of a $GP$ with $a$ as the first term and $r$ as the common ratio,then $S_n : S_{2n}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module