⚡ AZUCATION SHORTCUT:
Ratio of $\frac{a_1 a_2 + a_2 a_3 + a_3 a_1}{a_1 + a_2 + a_3} = \frac{624}{52} = 12$.
For a GP, this ratio is simply $a_2$. So, $ar = 12$.
Substitute $a = 12/r$ in sum: $\frac{12}{r} + 12 + 12r = 52 \implies 3r^2 - 10r + 3 = 0$.
Roots are $3$ and $1/3$. Since it is decreasing, $r = 1/3, a = 36$.
Sum $= \frac{36}{1 - 1/3} = \mathbf{54}$.

Step 1: Set up the equations
Let the first three terms be $a, ar, ar^2$.
1) $a + ar + ar^2 = a(1 + r + r^2) = 52$
2) $(a)(ar) + (ar)(ar^2) + (ar^2)(a) = a^2r + a^2r^3 + a^2r^2 = a^2r(1 + r^2 + r) = 624$

Step 2: Solve for 'ar' (the second term)
Divide Equation (2) by Equation (1):
$\frac{a^2r(1 + r + r^2)}{a(1 + r + r^2)} = \frac{624}{52}$
$ar = 12$

Step 3: Find 'a' and 'r'
Substitute $a = \frac{12}{r}$ into Equation (1):
$\frac{12}{r} + 12 + 12r = 52 \implies \frac{12}{r} + 12r = 40$
Divide by 4: $\frac{3}{r} + 3r = 10 \implies 3r^2 - 10r + 3 = 0$
Solving the quadratic: $(3r - 1)(r - 3) = 0$.
Since the GP is decreasing, $|r| < 1$, so $r = \frac{1}{3}$.
Then $a = \frac{12}{1/3} = 36$.

Step 4: Calculate Infinite Sum
Sum $S_{\infty} = \frac{a}{1 - r} = \frac{36}{1 - 1/3} = \frac{36}{2/3} = 36 \times \frac{3}{2} = \mathbf{54}$.

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