5th Term of Infinite Geometric Progression

Question

An infinite geometric progression $a_1, a_2, a_3, \dots$ has the property that $a_n = 3(a_{n+1} + a_{n+2} + \dots)$ for every $n \ge 1$. If the sum $a_1 + a_2 + a_3 + \dots = 32$, then $a_5$ is

Accepted Answer

C. Let the first term be $a$ and common ratio be $r$. Given $a_n = 3 	imes (	ext{sum of terms after } a_n)$: $$a_1 = 3(a_2 + a_3 + \dots)$$ $$a_1 = 3 \left(\frac{a_1 r}{1 - r}ight)$$ $$1 = \frac{3r}{1 - r} \implies 1 - r = 3r \implies 4r = 1 \implies r = \frac{1}{4}$$ Total sum = $\frac{a_1}{1 - r} = 32 \implies \frac{a_1}{3/4} = 32 \implies a_1 = 24$. 5th term $a_5 = a_1 r^4 = 24 	imes \left(\frac{1}{4}ight)^4 = 24 	imes \frac{1}{256} = \frac{3}{32}$.

Answer

1/32

Answer

2/32

Answer

3/32

Answer

4/32

5th Term of Infinite Geometric Progression

Explanation

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