First slide

Arithmetico-Geometric Progression

Question

$i f 3 + \frac{1}{4} (3 + P) + \frac{1}{4^{2}} (3 + 2 P) + \frac{1}{4^{3}} (3 + 3 P) + .... \infty = 8$ then the value of P is

Moderate

A

a prime number
B

an even number
C

is a perfect square of a natural number
D

is an irrational number

Solution

$T h e g i v e n s e r i e s c a n b e w r i t t e n a s 3 (1 + \frac{1}{4} + \frac{1}{4^{2}} + ......) + \frac{P}{4} (1 + 2. \frac{1}{4} + 3. \frac{1}{4^{2}} + .....) = 8$
Use the sum of infinte GP and infinite AGP formulae
$\begin{array}{rcl} \Rightarrow & 3 \times \frac{1}{1 - \frac{1}{4}} + \frac{P}{4} [\frac{1}{1 - \frac{1}{4}} + \frac{\frac{1}{4}}{{(1 - \frac{1}{4})}^{2}}] = 8 \\ \Rightarrow & 3 \times \frac{4}{3} + \frac{P}{4} (\frac{4}{3} + \frac{1}{4} \times \frac{16}{9}) = 8 \\ 4 + P (\frac{4}{9}) & = & 8 \\ P & = & 9 \end{array}$

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