Question

If \[3 + 3\alpha + 3{\alpha ^2} + ....\infty = \dfrac{{45}}{8}\], \[\alpha > 0\]. Then what is the value of \[\alpha \]?
A. \[\dfrac{{15}}{{23}}\]
B. \[\dfrac{7}{{15}}\]
C. \[\dfrac{7}{8}\]
D. \[\dfrac{{15}}{7}\]

Answer 1

Hint: The given terms of the given sum are in geometric progression. Use the formula of the sum of infinite terms in geometric progression and get the value of \[\alpha \].

Formula used:
Sum of infinite terms in geometric progression is: \[{S_\infty } = \dfrac{a}{{1 - r}}, - 1 < r < 1\].
Where \[a\] is the first term and \[r\] is the common ratio of a geometric progression.
The common ratio of a geometric progression \[a,ar,a{r^2},a{r^3},....\] is: \[r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}} = ...\]

Complete step by step solution:
The given sum is \[3 + 3\alpha + 3{\alpha ^2} + ....\infty = \dfrac{{45}}{8}\], \[\alpha > 0\].
Clearly, the terms of the equation are in geometric progression.
The common ratio of the given terms is \[r = \dfrac{{3\alpha }}{3} = \alpha \] and the first term is \[3\].
Now apply the sum of infinite terms in geometric progression.
\[3 + 3\alpha + 3{\alpha ^2} + ....\infty = \dfrac{3}{{1 - \alpha }}\]
\[ \Rightarrow \]\[\dfrac{{45}}{8} = \dfrac{3}{{1 - \alpha }}\]
Cross multiplies the terms of above equation.
\[45\left( {1 - \alpha } \right) = 3\left( 8 \right)\]
\[ \Rightarrow \]\[45 - 45\alpha = 24\]
\[ \Rightarrow \]\[45\alpha = 45 - 24\]
\[ \Rightarrow \]\[45\alpha = 21\]
Divide both sides by \[45\].
\[\alpha = \dfrac{{21}}{{45}}\]
\[ \Rightarrow \]\[\alpha = \dfrac{7}{{15}}\]
Hence the correct option is B.

Note: The sum of infinite geometry progress applicable when the common difference lies in the interval \[\left( { - 1,1} \right)\]. In the given question the common difference is \[\alpha \]. So the value of \[\alpha \] must be lie in the interval \[\left( { - 1,1} \right)\]. We know, \[\dfrac{7}{{15}} \in \left( { - 1,1} \right)\]. Hence we get correct value of \[\alpha \].