IPMAT Indore 2024 · Quantitative Ability · Question 31

Step 1 — spot the warning sign. The infinite sum is 80, yet we are asked when a partial sum is near 100. A partial sum can only exceed the limit if the terms alternate in sign, so expect r<0.

Step 2 — set up.

\frac{a}{1-r}=80, a+ar=a\left( 1+r \right)=35

From the first, a=80\left( 1-r \right). Substituting:

80\left( 1-r \right)\left( 1+r \right)=35 \Rightarrow 80\left( 1-{r}^{2}\right)=35 \Rightarrow 1-{r}^{2}=\frac{7}{16}

{r}^{2}=\frac{9}{16}\Rightarrow r=\pm \frac{3}{4}

Step 3 — reject r=+3/4. It gives a=20 and {S}_{n}=80\left( 1-{\left( 3/4 \right)}^{n}\right)<80 always — never close to 100. So

r=-\frac{3}{4}, a=80\left( 1+3/4 \right)=140

Step 4 — write the partial sums.

{S}_{n}=\frac{a\left( 1-{r}^{n}\right)}{1-r}=\frac{140\left( 1-{\left( -3/4 \right)}^{n}\right)}{7/4}=80\left( 1-{\left( -3/4 \right)}^{n}\right)

Step 5 — tabulate. Odd n overshoots, even n undershoots.

Answer: Option 2 — n=5

Shortcut. Let the structure choose the sign. “Partial sum near 100 when the infinite sum is 80” can only happen with r<0 — that single observation removes half the work.

After that, {S}_{n}=80\left( 1-{\left( -3/4 \right)}^{n}\right) means the partial sums oscillate around 80, overshooting on odd n and converging. Since you need a value near 100, only odd n qualifies: test n=3,5,7 and nothing else. n=5 lands at 98.98 — within 1. Three evaluations instead of seven.