UPSC NDA I 2019 Mathematics Solved Paper

Concept:A set of three numbers can be terms of infinitely many arithmetic progressions or geometric progressions if the indices can be chosen arbitrarily to satisfy the common difference or common ratio.Explanation:Let $1,5,25$ be the $p$-th, $q$-th, and $r$-th terms of an AP with first term $a$ and common difference $d$.Then:$a + (p-1)d = 1$$a + (q-1)d = 5$$a + (r-1)d = 25$Subtracting the first from the second gives $(q-p)d = 4$.Subtracting the second from the third gives $(r-q)d = 20$.Dividing these yields $\frac{r-q}{q-p}=5$, which is a rational number.Since $p,q,r$ can be chosen as any integers satisfying this ratio, infinitely many APs exist.Thus options (A) and (B) are incorrect.Now assume $1,5,25$ are the $P$-th, $Q$-th, and $R$-th terms of a GP with first term $a$ and common ratio $r$.Then:$a r^{P-1} = 1$$a r^{Q-1} = 5$$a r^{R-1} = 25$Dividing the second by the first gives $r^{Q-P}=5$.Dividing the third by the second gives $r^{R-Q}=5$.Thus $Q-P = R-Q$, i.e., $Q = \frac{P+R}{2}$.There are infinitely many choices for $P,Q,R$ satisfying this relation, each giving a valid GP.Therefore $1,5,25$ can be terms of infinitely many GPs.Answer:C. infinite number of GPs