Question

Let $a_1, a_2, a_3, ...$ be an A.P. If $\sum_{r=1}^{\infty} \frac{a_r}{2^r} = 4$, then $4a_2$ is equal to _____.

Answer 1

16

Answer 2

Let the arithmetic progression be denoted by $a_r = a + (r-1)d$, where $a = a_1$ is the first term and $d$ is the common difference. We are given the sum of an infinite series: $\sum_{r=1}^{\infty} \frac{a_r}{2^r} = 4$ Substituting the expression for $a_r$: $\sum_{r=1}^{\infty} \frac{a + (r-1)d}{2^r} = 4$ We can split this sum into two parts: $\sum_{r=1}^{\infty} \frac{a}{2^r} + \sum_{r=1}^{\infty} \frac{(r-1)d}{2^r} = 4$ $a \sum_{r=1}^{\infty} \left(\frac{1}{2}\right)^r + d \sum_{r=1}^{\infty} \frac{r-1}{2^r} = 4$ The first part is a geometric series with first term $1/2$ and common ratio $1/2$. The sum is: $\sum_{r=1}^{\infty} \left(\frac{1}{2}\right)^r = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$ So, the first term of the equation becomes $a \times 1 = a$.

The second part is $d \sum_{r=1}^{\infty} \frac{r-1}{2^r}$. Let's evaluate the sum $T = \sum_{r=1}^{\infty} \frac{r-1}{2^r}$: $T = \frac{1-1}{2^1} + \frac{2-1}{2^2} + \frac{3-1}{2^3} + \frac{4-1}{2^4} + \dots$ $T = \frac{0}{2} + \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots$ $T = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots$ This is an arithmetico-geometric series. Multiply $T$ by the common ratio $1/2$: $\frac{1}{2}T = \frac{1}{8} + \frac{2}{16} + \frac{3}{32} + \dots$ Subtracting $\frac{1}{2}T$ from $T$: $T - \frac{1}{2}T = \left(\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots\right) - \left(\frac{1}{8} + \frac{2}{16} + \frac{3}{32} + \dots\right)$ $\frac{1}{2}T = \frac{1}{4} + \left(\frac{2}{8} - \frac{1}{8}\right) + \left(\frac{3}{16} - \frac{2}{16}\right) + \left(\frac{4}{32} - \frac{3}{32}\right) + \dots$ $\frac{1}{2}T = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots$ This is a geometric series with first term $1/4$ and common ratio $1/2$. $\frac{1}{2}T = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$ So, $T = 1$.

Substituting the sums back into the equation: $a(1) + d(1) = 4$ $a + d = 4$ The second term of the AP is $a_2 = a + (2-1)d = a+d$. Therefore, $a_2 = 4$.

The question asks for the value of $4a_2$. $4a_2 = 4 \times 4 = 16$

Alternatively, using the formula for the sum of an arithmetico-geometric series $\sum_{n=1}^{\infty} [A + (n-1)D]x^{n-1} = \frac{A}{1-x} + \frac{Dx}{(1-x)^2}$ for $|x|<1$. The given sum is $S = \sum_{r=1}^{\infty} \frac{a + (r-1)d}{2^r} = \frac{1}{2} \sum_{r=1}^{\infty} [a + (r-1)d] \left(\frac{1}{2}\right)^{r-1}$. Here, $A=a$, $D=d$, and $x=1/2$. The sum inside the bracket is $\frac{a}{1-1/2} + \frac{d(1/2)}{(1-1/2)^2} = \frac{a}{1/2} + \frac{d/2}{(1/2)^2} = 2a + \frac{d/2}{1/4} = 2a + 2d$. So, $S = \frac{1}{2}(2a+2d) = a+d$. Given $S=4$, so $a+d=4$. Since $a_2 = a+d$, we have $a_2=4$. Thus, $4a_2 = 4 \times 4 = 16$.

The sum of the given arithmetico-geometric series $\sum_{r=1}^{\infty} \frac{a_r}{2^r}$ where $a_r$ is an AP, evaluates to $a_1 + d$, where $a_1$ is the first term and $d$ is the common difference of the AP. Given this sum is 4, we have $a_1 + d = 4$. The second term of the AP is $a_2 = a_1 + d$. Therefore, $a_2 = 4$. The value of $4a_2$ is $4 \times 4 = 16$.