Question

The sum of a given infinite geometric progression is 80 and the sum of its first two terms is 35. Then the value of \(n\) for which the sum of its first \(n\) terms is closest to 100, is:

• A. 6
• B. 5
• C. 7
• D. 4

Hint:

For geometric progressions, use the sum formula for the first \( n \) terms and the sum of an infinite series to derive relationships between terms and solve for unknowns.

Answer 1

The Correct Option is B

Given:

• Sum of infinite GP (\(S_\infty\)) = 80
• Sum of first two terms (\(S_2\)) = 35

Goal: Find \(n\) for which the sum of the first \(n\) terms (\(S_n\)) is closest to 100. Steps:

1. Formulate equations:

\[ S_\infty = \frac{a}{1 - r} = 80 \quad \text{(1)} \] \[ S_2 = a + ar = a(1 + r) = 35 \quad \text{(2)} \]

1. Solve for \(a\) and \(r\):

From (1), \(a = 80(1 - r)\).

Substitute in (2): \(80(1 - r)(1 + r) = 35\) \[ 80(1 - r^2) = 35 \] \[ 1 - r^2 = \frac{35}{80} = \frac{7}{16} \] \[ r^2 = 1 - \frac{7}{16} = \frac{9}{16} \] \[ r = \pm \frac{3}{4} \]

Since \(S_\infty\) exists, \(|r| < 1\). So \(r = \frac{3}{4}\) or \(r = -\frac{3}{4}\).

If \(r = \frac{3}{4}\), \(a = 80 \left( 1 - \frac{3}{4} \right) = 80 \left( \frac{1}{4} \right) = 20\).

If \(r = -\frac{3}{4}\), \(a = 80 \left( 1 + \frac{3}{4} \right) = 80 \left( \frac{7}{4} \right) = 140\).

1. Find \(S_n\): \[ S_n = \frac{a(1 - r^n)}{1 - r} \]
2. Test values of \(n\):
   • Case 1: \(a = 20\), \(r = \frac{3}{4}\) \[ S_n = \frac{20 \left( 1 - \left( \frac{3}{4} \right)^n \right)}{1 - \frac{3}{4}} = 80 \left( 1 - \left( \frac{3}{4} \right)^n \right) \] We want \(S_n \approx 100\). Then \[ 100 = 80 \left( 1 - \left( \frac{3}{4} \right)^n \right) \] \[ \frac{100}{80} = 1 - \left( \frac{3}{4} \right)^n \] \[ \frac{5}{4} = 1 - \left( \frac{3}{4} \right)^n \] \[ \left( \frac{3}{4} \right)^n = 1 - \frac{5}{4} = -\frac{1}{4} \quad \text{(Not possible)} \]
   • Case 2: \(a = 140\), \(r = -\frac{3}{4}\) \[ S_n = \frac{140 \left( 1 - \left( -\frac{3}{4} \right)^n \right)}{1 - \left( -\frac{3}{4} \right)} = \frac{140 \left( 1 - \left( -\frac{3}{4} \right)^n \right)}{\frac{7}{4}} = 80 \left( 1 - \left( -\frac{3}{4} \right)^n \right) \] We want \(S_n \approx 100\). Then \[ 100 = 80 \left( 1 - \left( -\frac{3}{4} \right)^n \right) \] \[ \frac{100}{80} = 1 - \left( -\frac{3}{4} \right)^n \] \[ \frac{5}{4} = 1 - \left( -\frac{3}{4} \right)^n \] \[ \left( -\frac{3}{4} \right)^n = 1 - \frac{5}{4} = -\frac{1}{4} \] Try \(n = 5\): \(\left( -\frac{3}{4} \right)^5 = -\frac{243}{1024} \approx -0.237\). Try \(n = 6\): \(\left( -\frac{3}{4} \right)^6 = \frac{729}{4096} \approx 0.178\). Try \(n = 4\): \(\left( -\frac{3}{4} \right)^4 = \frac{81}{256} \approx 0.316\). Then \[ 80 \left( 1 - \left( -\frac{3}{4} \right)^5 \right) \approx 80 (1 + 0.237) \approx 80 (1.237) \approx 98.96 \approx 99 \] \[ 80 \left( 1 - \left( -\frac{3}{4} \right)^6 \right) \approx 80 (1 - 0.178) \approx 80 (0.822) \approx 65.76 \] \[ 80 \left( 1 - \left( -\frac{3}{4} \right)^4 \right) \approx 80 (1 - 0.316) \approx 80 (0.684) \approx 54.72 \] Thus, \(S_n\) is closest to 100 when \(n = 5\)

Answer: (b) 5