Question

Let $a_1, a_2, a_3,...$ be a G.P. of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $a_1a_9 + a_2a_9 + a_5 + a_7$ is equal to

Answer 1

33 + 9\sqrt{7}

Answer 2

Let the first term of the Geometric Progression (GP) be $a$ and the common ratio be $r$. Since the GP consists of increasing positive numbers, we have $a > 0$ and $r > 1$. The $n$-th term of a GP is given by $a_n = a \cdot r^{n-1}$.

We are given two conditions:

1. The product of the fourth and sixth terms is 9: $a_4 \cdot a_6 = 9$ $(a \cdot r^{4-1}) \cdot (a \cdot r^{6-1}) = 9$ $(a \cdot r^3) \cdot (a \cdot r^5) = 9$ $a^2 \cdot r^8 = 9$ This can be written as $(a \cdot r^4)^2 = 9$. Since $a > 0$ and $r > 1$, $a \cdot r^4$ must be positive. Thus, $a \cdot r^4 = 3$. Note that $a_5 = a \cdot r^{5-1} = a \cdot r^4$. So, $a_5 = 3$.

2. The sum of the fifth and seventh terms is 24: $a_5 + a_7 = 24$ Since $a_5 = 3$, we have: $3 + a_7 = 24$ $a_7 = 21$.

Now we can find the common ratio $r$. $a_7 = a_5 \cdot r^{7-5}$ $21 = 3 \cdot r^2$ $r^2 = \frac{21}{3} = 7$ Since $r > 1$, we take the positive square root: $r = \sqrt{7}$.

We need to find the value of the expression $a_1a_9 + a_2a_9 + a_5 + a_7$. We already know $a_5 + a_7 = 24$. Now let's calculate the terms $a_1a_9$ and $a_2a_9$.

$a_1a_9 = a \cdot (a \cdot r^{9-1}) = a \cdot (a \cdot r^8) = a^2 r^8$. From the first condition, we found $a^2 r^8 = 9$. So, $a_1a_9 = 9$.

$a_2a_9 = (a \cdot r^{2-1}) \cdot (a \cdot r^{9-1}) = (a \cdot r) \cdot (a \cdot r^8) = a^2 r^9$. We can write $a^2 r^9$ as $(a^2 r^8) \cdot r$. Since $a^2 r^8 = 9$ and $r = \sqrt{7}$: $a_2a_9 = 9 \cdot \sqrt{7}$.

Now, substitute these values into the expression: $a_1a_9 + a_2a_9 + a_5 + a_7 = 9 + 9\sqrt{7} + 24$ $= 33 + 9\sqrt{7}$.