Verified answer

Suppose they get a, a+d, a+2d, a+3d, a+4d loaves.

So 5a+10d = 100 or a + 2d = 20..................(1)

From the given condition,

a+(a+d) = (1/7)[(a+2d)+ (a+3d) + (a+4d)]

which means 7[2a +d] = [3a+9d]....................(2)

Solve these 2 eqns simultaneouly to find a and d.

You should have said "The first two together receive one-seventh of what the last three receive TOGETHER"

let n represent the number of loaves given to the first person and m the value of the term to term difference. The the 5 people receive, in order, n, n+m, n+2m, n+3m, and n+4m loaves.

Adding, we get 5n + 10m = 100

and 7(2n +m) = 3n + 9m 14n + 7m = 3n + 9m

Solving we get 11n = 2m or 11n - 2m =0

solving this simultaneously with the first equation yields 60n=100 or n = 5/3

which yields m = 55/6

So they receive, in order 5/3, 65/6, 120/6 (=20), 175/6, and 230/6 (=115/3) loaves

Let the five receive the portions in teh following AP

(a - 2d), (a - d), a , (a + d), (a + 2d)

Then

(a - 2d) + (a - d) + a + (a + d) + (a + 2d) = 100

=> 5a = 100

=> a = 20

Now

(a - 2d) + (a - d) = 1/7 [a + (a + d) + (a + 2d)]

=> 11a = 24d

=> d = 11a/24

=> d = 9.17

Hence portions should be

(20 - 2*9.17), (20 - 9.17), 20 , (20 + 9.17), (20 + 2*9.17)

= 1.67, 10.83, 20, 29.17, 38.34