Let S be the sum of the first 9 terms of the series :
{x + k$a$} + {x² + (k + 2)$a$} + {x³ + (k + 4)$a$}
+ {x⁴ + (k + 6)$a$} + .... where a $\ne$ 0 and x $\ne$ 1.

If S = ${{{x^{10}} - x + 45a\left( {x - 1} \right)} \over {x - 1}}$, then k is equal to :

S = {x + k$a$} + {x² + (k + 2)$a$} + {x³ + (k + 4)$a$} <br>+ {x⁴ + (k + 6)$a$} + ....<br><br> $S = \left( {x + {x^2} + {x^3} + ....\,9terms} \right) +$<br>&nbsp;&nbsp;&nbsp;$a\left( {k + \left( {k + 2} \right) + (k + 4) + (k + 6) + ....9terms} \right)$<br><br> $$S = {{x\left( {{x^9} - 1} \right)} \over {\left( {x - 1} \right)}} + a\left[ {{9 \over 2}\left[ {2k + \left( {9 - 1} \right)2} \right]} \right]$$<br><br> $S = {{{x^{10}} - x} \over {x - 1}} + 9a\left( {k + 8} \right)$<br><br> $$S = {{{x^{10}} - x + 9a\left( {k + 8} \right)\left( {x - 1} \right)} \over {x - 1}}$$<br><br> $$S = {{{x^{10}} - x + 9\left( {k + 8} \right)a\left( {x - 1} \right)} \over {x - 1}}$$<br><br> Compare with given sum, then we get<br><br> ${9\left( {k + 8} \right) = 45}$<br><br> $\Rightarrow \left( {k + 8} \right) = 5$<br><br> $\Rightarrow k = - 3$