SolveIt is under development
SolveItJEE Main
Question 237 / 251 · latest first
Mathematics · 2020

JEE Main · 2 September 2020, Shift 2 · Q58

Let S be the sum of the first 9 terms of the series: {x+ ka }+{x^2+( k +2) a }+{x^3+( k +4) a } +{x^4+( k +6) a }+… where a ≠ 0 and x ≠ 1. If S…

Let $\displaystyle S$ be the sum of the first $\displaystyle 9$ terms of the series : $$\{x+\mathrm{ka}\}+\left\{x^2+(\mathrm{k}+2) \mathrm{a}\right\}+\left\{x^3+(\mathrm{k}+4) \mathrm{a}\right\} $$ $\displaystyle +\left\{x^4+(\mathrm{k}+6) \mathrm{a}\right\}+\ldots$ where $\displaystyle \mathrm{a} \neq 0$ and $\displaystyle x \neq 1$. If $\displaystyle \mathrm{S}=\frac{x^{10}-x+45 \mathrm{a}(x-1)}{x-1}$, then k is equal to :
ShareWhatsAppTelegram

More from Progressions

JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.