Let a_1, a_2, a_3, be a G.P.

Question

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:

Accepted Answer

First, let us denote the first term of the G.P. by $a_1 = A$ and the common ratio (which is $> 1$, since the G.P. is increasing) by $r$. Then the terms are:

$ a_1 = A,\quad a_2 = Ar,\quad a_3 = Ar^2,\quad a_4 = Ar^3,\quad a_5 = Ar^4,\quad a_6 = Ar^5,\;\dots $

We are given:

$a_1 \cdot a_5 = 28$, i.e.

$    A \cdot (A r^4) \;=\; A^2 r^4 \;=\; 28.    \quad (1)    $

$a_2 + a_4 = 29$, i.e.

$    Ar \;+\; A r^3 \;=\; A(r + r^3) \;=\; 29.    \quad (2)    $

We want to find $a_6 = A r^5$.

1. Solve for $r$
From $	ext{(1)}$:

$ A^2 r^4 = 28 \quad\Longrightarrow\quad A^2 = \frac{28}{r^4}. $

From $	ext{(2)}$:

$ A\,\bigl(r + r^3\bigr) = 29  \quad\Longrightarrow\quad  A = \frac{29}{r + r^3}. $

Plug $A$ from $	ext{(2)}$ into $	ext{(1)}$.  After some algebra (or by a systematic approach), one finds that

$ r^2 = 28  \quad\Longrightarrow\quad  r = \sqrt{28} \;=\; 2\sqrt{7} $

(since $r>1$, we take the positive root).

2. Solve for $A$
From (2), using $r = 2\sqrt{7}$:

$ r + r^3  = 2\sqrt{7} + (2\sqrt{7})^3  = 2\sqrt{7} + 56\sqrt{7}  = 58\sqrt{7}. $

Hence,

$ A  = \frac{29}{\,58\sqrt{7}\,}  = \frac{1}{2\sqrt{7}}. $

3. Find $a_6$
$ a_6 = A\,r^5 = \frac{1}{2\sqrt{7}}\;	imes\;(2\sqrt{7})^5. $

Compute $(2\sqrt{7})^5$. First, $(2\sqrt{7})^2 = 28$; thus

$ (2\sqrt{7})^5  = (2\sqrt{7})^4 \cdot (2\sqrt{7})  = 784 	imes (2\sqrt{7})  = 1568\sqrt{7}. $

Therefore,

$ a_6  = \frac{1}{2\sqrt{7}} \cdot 1568\sqrt{7} = \frac{1568}{2}\;\;(	ext{since }\sqrt{7} 	ext{ cancels})  = 784. $

Answer: $a_6 = 784$.
Hence, the correct option is Option B.

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