Question

Let $a_n$ be the $n^{\text{th}}$ term of a decreasing infinite geometric progression. If $a_1 + a_2 + a_3 = 52$ and $a_1a_2 + a_2a_3 + a_3a_1 = 624$, then the sum of this geometric progression is:

• A. \(57\)
• B. \(63\)
• C. \(54\)
• D. \(60\)

Hint:

For an infinite geometric progression with first term $a$ and common ratio $r$ (where $\lvert r \rvert<1$): \[ S_\infty = \frac{a}{1 - r}. \] Also, using relationships between sums and products of initial terms can help form equations in $a$ and $r$.

Answer 1

The Correct Option is C

Approach: The two given symmetric expressions in \(a_1,a_2,a_3\) are really expressions in the first term \(a\) and ratio \(r\). Write both, and dividing one by the other kills the messy \((1+r+r^2)\) factor in one stroke.

Step 1: Let the first term be \(a\) and common ratio be \(r\). Then \(a_1=a,\ a_2=ar,\ a_3=ar^2\), and from the first condition \[ a_1+a_2+a_3 = a(1+r+r^2) = 52. \tag{1} \]

Step 2: For the second condition, \[ a_1a_2+a_2a_3+a_3a_1 = a\cdot ar + ar\cdot ar^2 + ar^2\cdot a = a^2 r(1+r+r^2) = 624. \tag{2} \] Now notice the trick: equation (2) is just \(\big[a(1+r+r^2)\big]\times (a r) = 52\cdot(ar)\). So \[ 52\,(ar) = 624 \;\Rightarrow\; ar = 12. \tag{3} \] That single division removes the cubic clutter completely.

Step 3: Solve for \(r\). From (1), \(a+ar+ar^2 = 52\). Using \(ar=12\) and writing \(ar^2 = (ar)\cdot r = 12r\), we get \[ a + 12 + 12r = 52 \;\Rightarrow\; a = 40-12r. \] Multiply \(ar=12\): \((40-12r)r = 12 \Rightarrow 12r^2 - 40r + 12 = 0 \Rightarrow 3r^2 - 10r + 3 = 0.\) \[ r = \frac{10\pm\sqrt{100-36}}{6} = \frac{10\pm 8}{6} \;\Rightarrow\; r = 3 \ \text{or}\ r = \tfrac{1}{3}. \]

Step 4: Pick the valid root. The G.P. is decreasing and infinite, so we need \(|r|<1\); hence \(r=\tfrac{1}{3}\). Then \(a = \dfrac{12}{r} = 36.\)

Step 5: Sum to infinity: \[ S_\infty = \frac{a}{1-r} = \frac{36}{1-\tfrac{1}{3}} = \frac{36}{2/3} = 54. \] Final answer: \(54\).

Answer 2

Step 1: Assume the geometric progression. Let the first term be \(a\) and the common ratio be \(r\). Then the first three terms are \[ a_1 = a,\quad a_2 = ar,\quad a_3 = ar^2. \] Step 2: Use the given condition involving products. It is given that \[ a_1a_2 + a_2a_3 + a_3a_1 = 624. \] Substituting the terms of the G.P., \[ a(ar) + (ar)(ar^2) + (ar^2)(a) = a^2(r + r^3 + r^2). \] Rearranging, \[ a^2(r + r^2 + r^3) = a^2 r(1 + r + r^2) = 624. \tag{2} \] From the earlier condition (given in the question), \[ a(1 + r + r^2) = 52, \] so \[ a = \frac{52}{1 + r + r^2}. \tag{1} \] Substituting this value of \(a\) into equation (2), \[ \left(\frac{52}{1 + r + r^2}\right)^2 r(1 + r + r^2) = 624. \] Simplifying, \[ \frac{52^2 r}{1 + r + r^2} = 624. \] Hence, \[ 1 + r + r^2 = \frac{52^2 r}{624} = \frac{2704r}{624} = \frac{13}{3}r. \] Step 3: Solve for the common ratio. \[ 1 + r + r^2 = \frac{13}{3}r \Rightarrow 3 + 3r + 3r^2 = 13r \Rightarrow 3r^2 - 10r + 3 = 0. \] Solving this quadratic, \[ r = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm 8}{6}. \] Thus, \[ r = 3 \quad \text{or} \quad r = \frac{1}{3}. \] Since the geometric progression is decreasing and infinite, \(|r| < 1\). Therefore, \[ r = \frac{1}{3}. \] Step 4: Find the first term. \[ 1 + r + r^2 = 1 + \frac{1}{3} + \frac{1}{9} = \frac{13}{9}. \] From equation (1), \[ a = \frac{52}{13/9} = 36. \] Step 5: Find the sum of the infinite G.P. \[ S_\infty = \frac{a}{1 - r} = \frac{36}{1 - \frac{1}{3}} = \frac{36}{\frac{2}{3}} = 54. \] Therefore, the sum of the given geometric progression is \[ 54. \]