Question

Let $a_1, a_2, a_3,...$ be a G.P. of increasing positive terms. If $a_1a_3 = 28$ and $a_2 + a_4 = 29$, then $a_6$ is equal to:

Answer 1

$\frac{869\sqrt7 - 812}{14}$

Answer 2

We start by writing the terms of the GP as

$a_1 = a, a_2 = ar, a_3 = ar^2, a_4 = ar^3, \ldots$

Step 1. From the condition

$a_1 a_3 = a \cdot (ar^2) = a^2 r^2 = 28$,

we get

(ar)^2 = 28 \quad\Longrightarrow\quad ar = 2\sqrt7 \quad \text{(since a>0,, r>0)}.

Step 2. The second given condition is

$a_2 + a_4 = ar + ar^3 = ar(1 + r^2) = 29$.

Using $ar=2\sqrt7$ from Step 1, we have

$2\sqrt7 (1 + r^2) = 29 \quad\Longrightarrow\quad 1 + r^2 = \frac{29}{2\sqrt7}$.

Thus,

$r^2 = \frac{29}{2\sqrt7} - 1 = \frac{29 - 2\sqrt7}{2\sqrt7}$.

Step 3. We need to compute the sixth term:

$a_6 = ar^5 = (ar)\,r^4 = 2\sqrt7 \cdot r^4$.

But $r^4 = (r^2)^2$. So,

$r^4 = \left(\frac{29-2\sqrt7}{2\sqrt7}\right)^2 = \frac{(29-2\sqrt7)^2}{4\cdot7} = \frac{(29-2\sqrt7)^2}{28}$.

Thus,

$a_6 = 2\sqrt7 \cdot \frac{(29-2\sqrt7)^2}{28} = \frac{\sqrt7\,(29-2\sqrt7)^2}{14}$.

Step 4. To simplify further, expand $(29-2\sqrt7)^2$:

$(29-2\sqrt7)^2 = 29^2 - 2\cdot 29\cdot2\sqrt7 + (2\sqrt7)^2 = 841 - 116\sqrt7 + 28 = 869 - 116\sqrt7$.

Therefore,

$a_6 = \frac{\sqrt7\,(869-116\sqrt7)}{14} = \frac{869\sqrt7 - 116\cdot7}{14} = \frac{869\sqrt7 - 812}{14}$.