Mathematics - Sequence & Series Question with Solution | TestHub

If $a,b,c$ are in H.P., then $1/a,1/b,1/c$ are in A.P.

Given $a=3,b=4$, so $1/a=1/3,1/b=1/4$.

The common difference $d^{\prime }$ of the A.P. is $1/b-1/a=1/4-1/3=(3-4)/12=-1/12$. So

(Q)-(4). For the term $c$, $1/c=1/b+d^{\prime }=1/4+(-1/12)=3/12-1/12=2/12=1/6$.

Thus $c=6$. So (P)-(2).$1/a+1/c=1/3+1/6=2/6+1/6=3/6=1/2$. So

(R)-(1).$a+b+c=3+4+6=13$. So

(S)-(3).