IPMAT Indore 2025 · Quantitative Ability · Question 33

Step 1 — turn the equal-sums condition into something usable.

{S}_{7}={S}_{5}\Rightarrow{S}_{7}-{S}_{5}=0 \Rightarrow{t}_{6}+{t}_{7}=0

Step 2 — write those terms and factor.

a{r}^{5}+a{r}^{6}=0 \Rightarrow a{r}^{5}\left( 1+r \right)=0

Since a\ne 0 and r\ne 0 (otherwise there is no GP), we need

r=-1

Step 3 — what r=-1 does to the sums. The GP is a,-a,a,-a,\ldots — consecutive terms cancel in pairs. So:

{S}_{\text{even}}=0

{S}_{\text{odd}}=a

Step 4 — use {S}_{9}=24. Nine is odd, so {S}_{9}=a=24.

Step 5 — the 4th term.

{t}_{4}=a{r}^{3}=24\times{\left( -1 \right)}^{3}=-24

Answer: Option 2 — -24