XAT 2010 Question 82

Question 82

a,  b,  c,  d and  e are integers such that 1 ≤ 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 the maximum value of $$\frac{1}{lcm(a,b)}+\frac{1}{lcm(b,c)}+\frac{1}{lcm(c,d)}+\frac{1}{lcm(d,e)}$$ is

Solution

Given that the numbers are in G.P.

Let the common ratio be ‘r’, hence the series a,b,c,d,e can also be expressed as:

$$a , ar , ar^2 , ar^3 , ar^4$$

lcm(a,b) = lcm$$(a,ar) = ar$$

lcm(b,c) = lcm$$(ar,ar^2) = ar^2$$

lcm(c,d) = lcm$$(ar^2,ar^3) = ar^3$$

lcm(d,e) = lcm$$(ar^3,ar^4) = ar^4$$

$$\therefore \frac{1}{lcm(a,b)}+\frac{1}{lcm(b,c)}+\frac{1}{lcm(c,d)}+\frac{1}{lcm(d,e)}$$

= $$\frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} + \frac{1}{ar^4}$$

= $$\frac{1}{a} (\frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4})$$

To get max value of this, ‘a’ and ‘r’ should be minimum.

It is given that $$1 \leq a$$ => Minimum value of ‘a’ = 1

For the values in the series to be integers, the minimum common ratio, r = 2   ($$r \leq 1$$ won’t work here as it is an increasing GP)

Substituting values of 'a' and 'r' in the expression, we get :

Max value = $$\frac{1}{1} (\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4})$$

= $$\frac{8 + 4 + 2 + 1}{16} = \frac{15}{16}$$



Create a FREE account and get:

  • All Quant Formulas and shortcuts PDF
  • 40+ previous papers with solutions PDF
  • Top 500 MBA exam Solved Questions for Free

    Comments

    Register with

    OR

    Boost your Prep!

    Download App