IPMAT Indore 2025 — Question 14

Each term is $\ln a-k\cdot \ln b$ for an AP of $k$ starting at 1 with common difference $1/2$. Sum of 21 terms: $21\ln a-\ln b\cdot {\sum }_{i=1}^{21}(1+(i-1)/2)$. The exponent sum for $b$ is $\sum (1,1.5,2,2.5,\dots )=21+0.5\cdot \left(\genfrac{}{}{0ex}{}{21}{2}\right)/2$, simplifying to 126. So $m=21$, $n=126$, $m+n=147$.