IPMAT Indore 2025 · Quantitative Ability · Question 14

Step 1 — use the AP condition. If three numbers are in AP, the middle one is the average of the outer two, i.e. 2{t}_{2}={t}_{1}+{t}_{3}:

2{\log}_{3}\left( 2{x}^{2}+1 \right)={\log}_{3}\left({x}^{2}-1 \right)+{\log}_{3}\left( 6{x}^{2}+3 \right)

Step 2 — collapse the logs. Using 2\log M=\log{M}^{2} and \log P+\log Q=\log PQ:

{\left( 2{x}^{2}+1 \right)}^{2}=\left({x}^{2}-1 \right)\left( 6{x}^{2}+3 \right)=3\left({x}^{2}-1 \right)\left( 2{x}^{2}+1 \right)

Step 3 — cancel the common factor. Write u=2{x}^{2}+1 (note u\ne 0):

{u}^{2}=3\left({x}^{2}-1 \right)\,u \Rightarrow u=3\left({x}^{2}-1 \right)

2{x}^{2}+1=3{x}^{2}-3 \Rightarrow{x}^{2}=4

Step 4 — find the actual terms. With {x}^{2}=4:

{t}_{1}={\log}_{3}3=1,{t}_{2}={\log}_{3}9=2,{t}_{3}={\log}_{3}27=3

So the AP is 1,2,3,\ldots with common difference 1.

Step 5 — the next three terms are 4, 5, 6:

4+5+6=15

Answer: 15