JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Sequences and Series

From Vieta's formulas: $p+r=8a$, $pr=2a$, so $\frac{1}{p}+\frac{1}{r}=\frac{p+r}{pr}=\frac{8a}{2a}=4$. Similarly, $q+s=-12b$, $qs=6b$, so $\frac{1}{q}+\frac{1}{s}=\frac{-12b}{6b}=-2$. Since $\frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{s}$ are in A.P., let the common difference be $d$. Then $\frac{1}{q}=\frac{1}{p}+d$, $\frac{1}{r}=\frac{1}{p}+2d$, $\frac{1}{s}=\frac{1}{p}+3d$. From $\frac{1}{p}+\frac{1}{r}=4$: $2\cdot \frac{1}{p}+2d=4$, so $\frac{1}{p}+d=2$, i.e., $\frac{1}{q}=2$, giving $q=\frac{1}{2}$. From $\frac{1}{q}+\frac{1}{s}=-2$: $2+\frac{1}{s}=-2$, so $\frac{1}{s}=-4$, giving $s=-\frac{1}{4}$. Also, $\frac{1}{r}=2d+\frac{1}{p}=4-\frac{1}{p}$. From $\frac{1}{p}+\frac{1}{r}=4$, we get $\frac{2}{p}=4-2d$. Since $\frac{1}{q}=2$, we have $d=2-\frac{1}{p}$. Solving gives $\frac{1}{p}=5$, $p=\frac{1}{5}$, and $\frac{1}{r}=-1$, $r=-1$. Now, $\frac{1}{a}=\frac{2}{pr}=\frac{2}{(1/5)(-1)}=-10$, and $\frac{1}{b}=\frac{6}{qs}=\frac{6}{(1/2)(-1/4)}=-48$. Thus, $\frac{1}{a}-\frac{1}{b}=-10-(-48)=38$.