IIT-JEE 2012 · Paper 2 · Mathematics Q48

MathematicsSingle correct+3 / -1Question 48 of 60

Let $a_1,a_2,a_3,\ldots$ be in harmonic progression with $a_1=5$ and $a_{20}=25$ . The least positive integer $n$ for which $a_n<0$ is

Answer and worked solution

Official answer: D

Reciprocals form an AP. $\frac1{a_n}=\frac15+(n-1)d$ with
$\frac1{25}=\frac15+19d\implies d=-\frac4{475}.$

Negative terms. $a_n<0$ when $\frac15-\frac{4(n-1)}{475}<0$ , i.e. $n-1>\frac{475}{20}=23.75$ . The least such $n$ is 25.

(D)

Topics:Harmonic progression Arithmetic progression

IIT-JEE 2012 · Paper 2, Mathematics question 48