JEE Advanced 2023 Paper 1 · Q10 · Sequences & Series
Let $\underbrace{75\ldots57}_{r+2 \text{ digits}}$ denote the $(r+2)$ digit number where the first and the last digits are $7$ and the remaining $r$ digits are $5$. Consider the sum $S = 77 + 757 + 7557 + \ldots + \underbrace{75\ldots57}_{99 \text{ digits}}$. If $S = \dfrac{\underbrace{75\ldots57}_{99 \text{ digits}} + m}{n}$, where $m$ and $n$ are natural numbers less than $3000$, then the value of $m + n$ is
Reveal answer + step-by-step solution
Correct answer:1219
Solution
Write each term $77 = 7 \cdot 10 + 7$, $757 = 7\cdot 10^2 + 5\cdot 10 + 7$, generally $\underbrace{75\ldots57}_{r+2} = 7\cdot 10^{r+1} + 5(10^r + 10^{r-1}+\ldots+10) + 7 = 7\cdot 10^{r+1} + \dfrac{5(10^{r+1}-10)}{9} + 7$. Summing $r = 0$ to $97$: $S$ simplifies (geometric series) to $\dfrac{1}{9}(\underbrace{75\ldots57}_{99} + 1210)$, so $m = 1210$, $n = 9$, $m+n = 1219$.
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