Mathematics - Sequence & Series Question with Solution | TestHub

Item P: $p\ln a=q\ln b=r\ln c=\lambda$, so $\ln a,\ln b,\ln c=\frac{\lambda }{p},\frac{\lambda }{q},\frac{\lambda }{r}$. These are reciprocals of the G.P. $p,q,r$ (scaled), so they are in G.P. P matches 2. Item Q: $\frac{2a}{a-b{e}^x}-1=\frac{2b}{b-c{e}^x}-1=\frac{2c}{c-d{e}^x}-1$, so $\frac{a-b{e}^x}{a}=\frac{b-c{e}^x}{b}=\frac{c-d{e}^x}{c}$, giving $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$. a,b,c,d are in G.P. Q matches 2. Item R: $2b=a+c,p^2=ab,q^2=bc$, so $p^2+q^2=b(a+c)=2b^2$. In A.P. R matches 1. Item S: $b^2=ac$ gives $2{\log }_{10}b={\log }_{10}a+{\log }_{10}c$, i.e. $\frac{2}{{\log }_{b}10}=\frac{1}{{\log }_{a}10}+\frac{1}{{\log }_{c}10}$. In H.P. S matches 3.