Four distinct positive numbers, 'a', 'b', 'c' and 'd', in the order given, are in proportion. 'b' is 35 more than 'a' and 'd' is 60 more than 'c'. The product of 'a' and 'c' is 5376. What is the sum of 'a', 'b', 'c' and 'd'?

Given that a,b,c,d are in proportion:
b
a
​
=
d
c
​

We are given:

b=a+35

d=c+60

a×c=5376

Substitute the expressions for b and d into the proportion equation:

a+35
a
​
=
c+60
c
​

Cross-multiplying gives:

a(c+60)=c(a+35)
ac+60a=ac+35c
60a=35c
c
a
​
=
60
35
​
=
12
7
​

Let a=7k and c=12k for some positive constant k.
We know that a×c=5376:

(7k)×(12k)=5376
84k
2
=5376
k
2
=
84
5376
​
=64
k=
64

​
=8(since numbers are positive)
Now, find the values of a,b,c,d:

a=7×8=56
c=12×8=96
b=a+35=56+35=91
d=c+60=96+60=156
The sum of a,b,c,d is:

Sum=a+b+c+d=56+91+96+156=399