Sequences and Series
PYP_JEE_ADV_2023_P1
Grade None
Question:
Let $\overset{r}{75\cdots57}$ 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 $S=77+757+7557+\cdots+\overset{98}{75\cdots57}$. If $S=\dfrac{\overset{99}{75\cdots57}+m}{n}$, where $m$ and $n$ are natural numbers less than 3000, then the value of $m+n$ is ___.
Step-by-Step Solution
Key Concept: Express A_r = (680·10^r + 13)/9; sum the GP; then relate to A_99
For $r$ fives: $A_r=7\cdot10^{r+1}+5\cdot\dfrac{10^r-1}{9}\cdot10+7$... more cleanly:
$$A_r = \frac{680\cdot10^r+13}{9}.$$
Check: $A_0=(680+13)/9=693/9=77$ ✓, $A_1=(6800+13)/9=757$ ✓.
$S=\displaystyle\sum_{r=0}^{98}A_r=\dfrac{1}{9}\left(680\cdot\dfrac{10^{99}-1}{9}+13\cdot99\right)=\dfrac{680(10^{99}-1)+1287}{81}=\dfrac{680\cdot10^{99}+607}{81}$.
$A_{99}=\dfrac{680\cdot10^{99}+13}{9}$, so $9\cdot A_{99}=680\cdot10^{99}+13$.
$81S=680\cdot10^{99}+607=9\cdot A_{99}-13+607=9\cdot A_{99}+594$.
$S=\dfrac{9\cdot A_{99}+594}{81}=\dfrac{A_{99}+66}{9}$.
So $m=66$, $n=9$. $m+n=75$.
Correct Answer: 75