Question

a, b, c, d and e are integers such that $1\le a<b<c<d<e$. If a, b, c, d and e are geometric progression and lcm (m, n) is the least common multiple of m and n, then maximum value of
$\frac{1}{lcm(a,b)}+\frac{1}{lcm(b,c)}+\frac{1}{lcm(c,d)}+\frac{1}{lcm(d,e)}$ is

• A. $1$
• B. $\frac{15}{16}$
• C. $\frac{79}{81}$
• D. $\frac{7}{8}$

Answer 1

The correct option is B $\frac{15}{16}$

As a, b, c, d & e be in GP. Thus they can be expressed as $a,ax,a{x}^2,a{x}^3,a{x}^4$ where x is the common ratio of GP.
$\ell cm(a,b)=\ell cm(a,ax)=ax=bSimilarly,\ell cm(b,c)=c\ell cm(c,d)=dand\ell cm(d,e)=e$
Thus the expression
$\frac{1}{\ell cm(a,b)}+\frac{1}{\ell cm(b,c)}+\frac{1}{\ell cm(c,d)}+\frac{1}{\ell cm(d,e)}=\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}$
For the expression to be maximum, b,c,d and e should have minimum value. It is possible only when a is minimum i.e.. = 1
Thus the GP with integers having minimum value with first term = 1 will be 1, 2, 4, 8, 16.
$\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{8+4+2+1}{16}=\frac{15}{16}$