Let $a_{1}, a_{2}, a_{3}, \dots$ 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:
- (1) 628
- (2) 526
- (3) 784
- (4) 812
Solution:
Let the first term be $a$ and the common ratio be $r$. Since terms are increasing, $r > 1$.
Given $a_1 a_5 = 28 \Rightarrow a \cdot (ar^4) = 28 \Rightarrow a^2 r^4 = 28 \quad \dots(1)$
Given $a_2 + a_4 = 29 \Rightarrow ar + ar^3 = 29 \Rightarrow ar(1+r^2) = 29$
Squaring both sides:
$$ a^2 r^2 (1+r^2)^2 = (29)^2 $$
Substituting $a^2 = \frac{28}{r^4}$ from equation (1):
$$ \frac{28}{r^4} \cdot r^2 (1+r^2)^2 = 29^2 $$
$$ 28 \frac{(1+r^2)^2}{r^2} = 29^2 $$
Taking the square root on both sides:
$$ \sqrt{28} \frac{1+r^2}{r} = 29 \Rightarrow \frac{r}{1+r^2} = \frac{\sqrt{28}}{29} $$
Comparing the terms, we get $r = \sqrt{28}$.
Now, finding $a$ from equation (1):
$$ a^2 (\sqrt{28})^4 = 28 \Rightarrow a^2 (28)^2 = 28 \Rightarrow a^2 = \frac{1}{28} \Rightarrow a = \frac{1}{\sqrt{28}} $$
We need to find $a_6$:
$$ a_6 = ar^5 = \frac{1}{\sqrt{28}} (\sqrt{28})^5 = (\sqrt{28})^4 = (28)^2 = 784 $$
Ans. (3)