CAT 2025 Slot 3 · Quant · Question 2

In an AP the terms {t}_{4}, {t}_{7} and {t}_{10} are equally spaced about {t}_{7}, so their sum is 3{t}_{7}.

Hence 3{t}_{7}=99 and {t}_{7}=33.

The sum of the first 14 terms is \frac{14}{2}\left({t}_{1}+{t}_{14}\right), and {t}_{1}+{t}_{14}={t}_{7}+{t}_{8},

so {S}_{14}=7\left({t}_{7}+{t}_{8}\right).

So 7\left({t}_{7}+{t}_{8}\right)=497, giving {t}_{7}+{t}_{8}=71 and {t}_{8}=71-33=38.

The common difference is d={t}_{8}-{t}_{7}=38-33=5.

Then {t}_{1}={t}_{7}-6d=33-30=3.

Finally {S}_{5}=\frac{5}{2}\left( 2\times 3+4\times 5 \right)=\frac{5}{2}\times 26=65.

Shortcut. Any three terms symmetric about a middle term sum to three times that middle term, and any even-length sum is the count times the average of the two central terms. Those two facts give {t}_{7}=33 and {t}_{8}=38 immediately, and the whole AP follows from a single subtraction.