Let a_1, a_2, a_3, .

Question

Let $a_1, a_2, a_3, \ldots$. be a G.P. of increasing positive numbers. If $a_3 a_5=729$ and $a_2+a_4=\frac{111}{4}$, then $24\left(a_1+a_2+a_3\right)$ is equal to

Accepted Answer

We start with the given sequence $a_1, a_2, a_3, \ldots$ of an increasing geometric progression (G.P.). Two key conditions are provided:

$a_3 \cdot a_5 = 729$

$a_2 + a_4 = \frac{111}{4}$

Calculating using given conditions:

Using the sequence terms:

$    a_3 = a \cdot r^2, \quad a_5 = a \cdot r^4,     $

Then:

$    a_3 \cdot a_5 = (a \cdot r^2)(a \cdot r^4) = a^2 \cdot r^6 = 729 = 27^2    $

It follows:

$    a \cdot r^3 = 27 \quad \Rightarrow \quad a_4 = a \cdot r^3 = 27 \quad 	ext{(i)}    $

Using the second condition:

$    a_2 + a_4 = \frac{111}{4}    $

Substituting $a_4 = 27$:

$    a_2 = \frac{111}{4} - 27    $

$    a_2 = a \cdot r = \frac{3}{4} \quad 	ext{(ii)}    $

Solving for $r$ and $a$:

From equation (i) and (ii), we derive $r^2$:

$   r^2 = \frac{4 \cdot 27}{3} = 4 	imes 9 = 36   $

Since the G.P. is increasing, $r = 6$.

Solve for $a$:

Substituting $r = 6$ into $a \cdot r = \frac{3}{4}$:

$   a \cdot 6 = \frac{3}{4} \quad \Rightarrow \quad a = \frac{3}{24} = \frac{1}{8}   $

Calculating the requested sum:

The task is to find:

$ 24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) $

Substitute the values:

$ = 24 	imes \frac{1}{8}(1 + 6 + 6^2) = 3 	imes (1 + 6 + 36) = 3 	imes 43 = 129 $

Therefore, the value is 129.

Solutions · 1