Question

An infinite geometric progression a1 , a2 , a3 ,... has the property that an = 3(an+ l + an+2 +….) for every n ≥ 1. If the sum a1 + a2 + a3 +….... = 32, then a5 is

• A. 1/32
• B. 2/32
• C. 3/32
• D. 4/32

Answer 1

The Correct Option is C

An infinite geometric progression is given, and we know:

The condition: \(a_n = 3(a_{n+1} + a_{n+2} + \ldots)\) for every \(n \geq 1\).

The sum: \(a_1 + a_2 + a_3 + \ldots = 32\).

We need to find \(a_5\).

Given \(a_1, a_2, a_3, \ldots\) form a geometric progression, let the first term be \(a_1 = a\) and the common ratio be \(r\), so:

1. \[S = a + ar + ar^2 + \ldots = \frac{a}{1-r} = 32\] (sum of infinite GP)

2. From the condition, for \(n=1\):

\[a_1 = 3(a_2 + a_3 + \ldots) = 3(S - a_1)\]

3. Substituting \(S=32\):

\[a_1 = 3(32 - a_1)\]

4. Simplifying:

\[a_1 = 96 - 3a_1\]

\[4a_1 = 96\]

\[a_1 = 24\]

5. The sum of the series is:

\(\frac{a_1}{1-r} = 32\)

\(\frac{24}{1-r} = 32\)

6. Solving for \(r\):

\[24 = 32(1-r)\]

\[24 = 32 - 32r\]

\[32r = 8\]

\[r = \frac{1}{4}\]

7. \(a_5 = ar^4 = 24 \left(\frac{1}{4}\right)^4 = 24 \times \frac{1}{256} = \frac{24}{256} = \frac{3}{32}\)

Therefore, the value of \(a_5\) is \( \frac{3}{32} \).