Mathematics - Sequence & Series Question with Solution | TestHub

(A) $b+c,c+a,a+b$ are in H.P. $⟺\frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}⟺b^2-a^2=c^2-b^2⟺a^2,b^2,c^2$ are in A.P. (B) The terms are $A+(p-1)d,A+(q-1)d,\dots$, so differences of consecutive chosen terms are $(p-q)d,(q-$ $r)d,(r-s)d$. If the four terms are in G.P. with ratio $\rho$, then $a_q-a_r=\rho \left(a_p-a_q\right)$ and $a_r-a_s=\rho \left(a_q-a_r\right)$, so $p-q,q-$ $r,r-s$ are in G.P. (C) With $b=\frac{2ac}{a+c}:\frac{1}{b-a}+\frac{1}{b-c}=\frac{a+c}{ac-a^2}+\frac{a+c}{ac-c^2}=\frac{a+c}{ac}=\frac{1}{a}+\frac{1}{c}$. (D) is therefore false.