Question

If the sum of the infinity of the series $3 + 5r + 7{r^2} + ......$is $\dfrac{{44}}{9}$ , find the value of r.

Answer 1

Here we need to take help of geometric progression and infinite series together.
1. .Sn=$\left( {\dfrac{a}{{1 - r}}} \right)$
After forming a quadratic equation use the given formula for finding root.
2. $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step answer:
Let the sum given is denoted by letter s.
$s = 3 + 5r + 7{r^2} + ......$
$rs = 3r + 5{r^2} + 7{r^3} + ......$ multiplying both sides by r.
$s - rs = 3 + 2r + 2{r^2} + 2{r^3} + ....$ subtracting the two series.
$s(1 - r) = 3 + 2(r + {r^2} + {r^3} + .....)$ taking s common on left side and r common on right side

$$s(1 - r) = 1 + 2 + 2(r + {r^2} + {r^3} + .....) \\\ s(1 - r) = 1 + 2(1 + r + {r^2} + {r^3} + .....) \\\$$

Now the series $1 + r + {r^2} + {r^3} + .....$ is a geometric progression.
Thus, sum of terms in a G.P. is given by $s_n$=$\left( {\dfrac{a}{{1 - r}}} \right)$

$$\Rightarrow s(1 - r) = 1 + 2\left( {\dfrac{a}{{1 - r}}} \right) \\\ \Rightarrow s(1 - r) = 1 + 2\left( {\dfrac{1}{{1 - r}}} \right) \\\$$

$s(1 - r) = \dfrac{{1 - r + 2}}{{1 - r}}$ taking L.C.M.on right side
$s{(1 - r)^2} = 3 - r$
$s(1 - 2r + {r^2}) = 3 - r$
$\dfrac{{44}}{9}(1 - 2r + {r^2}) = 3 - r$ substitute value of s.

$$\Rightarrow 44(1 - 2r + {r^2}) = 9(3 - r) \\\ \Rightarrow 44 - 88r + 44{r^2} = 27 - 9r \\\ \Rightarrow 44{r^2} - 88r + 9r + 44 - 27 = 0 \\\ \Rightarrow 44{r^2} - 79r + 17 = 0 \\\$$

Now this equation is in quadratic equation form $a{x^2} + bx + c = 0$ having roots to be found using
formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Here a=44,b=-79 and c=17.putting the values,

$$\Rightarrow r = \dfrac{{ - ( - 79) \pm \sqrt {{{( - 79)}^2} - 4 \times 44 \times 17} }}{{2 \times 44}} \\\ \Rightarrow r = \dfrac{{79 \pm \sqrt {6241 - 2992} }}{{88}} \\\ \Rightarrow r = \dfrac{{79 \pm \sqrt {3249} }}{{88}} \\\ \Rightarrow r = \dfrac{{79 \pm 57}}{{88}} \\\ \Rightarrow r = \dfrac{{79 + 57}}{{88}},\dfrac{{79 - 57}}{{88}} \\\ \Rightarrow r = \dfrac{{136}}{{88}},\dfrac{{22}}{{88}} \\\ \Rightarrow r = \dfrac{{17}}{{11}},\dfrac{1}{4} \\\$$

Thus, we found two values of r=$\dfrac{{17}}{{11}},\dfrac{1}{4}$.

Note: Since the given series is not a G.P. we need to convert it using some mathematical operations
A geometric series is of the form $a + ar + a{r^2} + a{r^3} + ......$.