CAT 2025 Slot 2 · Quant · Question 8

Let the three terms be a, ar and a{r}^{2}.

The first condition gives a\left( 1+r+{r}^{2}\right)=52.

Expanding the second, {a}^{2}r+{a}^{2}{r}^{3}+{a}^{2}{r}^{2}={a}^{2}r\left( 1+r+{r}^{2}\right).

That is ar\times a\left( 1+r+{r}^{2}\right)=52\,ar,

so 52\,ar=624 and ar=12.

Substituting a=\frac{12}{r} into the first condition: \frac{12}{r}\left( 1+r+{r}^{2}\right)=52.

Clearing the fraction, 12{r}^{2}-40r+12=0, that is 3{r}^{2}-10r+3=0.

Factorising, \left( 3r-1 \right)\left( r-3 \right)=0, so r=\frac{1}{3} or r=3.

A decreasing infinite GP needs \left| r \right|<1, so r=\frac{1}{3} and a=36.

The sum to infinity is \frac{a}{1-r}=\frac{36}{\frac{2}{3}}=54.

Shortcut. The second expression always factors as (second term) \times (sum of the three terms) for three terms in GP. So {a}_{2}=\frac{624}{52}=12 in one division, and the rest is a single quadratic.