## Saturday, 29 October 2011

### Progressions-5 (XAT-2010)

a,b,c,d and e are integers such that 1 ≤ a < b < c < d < e. If a,b,c,d and e are in geometric progression and lcm(m,n) is the least common multiple of m and n, then the maximum value of 1/lcm(a,b) + 1/lcm(b,c) + 1/lcm(c,d) + 1/lcm(d,e) is:
(A)1      (B) 15/16        (C) 79/81        (D) 7/8   (E) None of these

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,ar2,ar3,ar4
lcm(a,b) = lcm(a,ar) = ar
lcm(b,c) = lcm(ar,ar2) = ar2
lcm(c,d) = lcm(ar2,ar3) = ar3
lcm(d,e) = lcm(ar3,ar4) = ar4
1/lcm(a,b) + 1/lcm(b,c) + 1/lcm(c,d) + 1/lcm(d,e)
= 1/ar + 1/ar2 + 1/ar3 + 1/ar4
= 1/a (1/r + 1/r2 + 1/r3 + 1/r4)                    --------(1)
To get max value of this, ‘a’ and ‘r’ should be minimum.
It is given that 1 ≤ a => Minimum value of ‘a’ = 1
For the values in the series to be integers, the minimum common ratio,
r = 2
(r ≤ 1 won’t work here as it is an increasing GP)
Substituting ‘a’ and ‘r’ values in the expression (1),
Maximum value = 1/2+ 1/4 + 1/8 + 1/16 = (8+4+2+1)/16 = 15/16