Question

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.

Answer 1

Given: Divide 56 into four parts in A.P. so that (product of extremes) : (product of means) = 5 : 6.

Step-wise calculation:

1. Let the four terms be in symmetric form about the mean:

14 – 3r, 14 – r, 14 + r, 14 + 3r

Since their sum = 56 ⇒ mean = `56/4` = 14.

2. Product of extremes = (14 – 3r)(14 + 3r) = 196 – 9r².

Product of means = (14 – r)(14 + r) = 196 – r².

3. Given (196 – 9r²) : (196 – r²) = 5 : 6 

⇒ `(196 - 9r^2)/(196 - r^2) = 5/6`

4. Cross-multiply and solve:

6(196 – 9r²) = 5(196 – r²)

1176 – 54r² = 980 – 5r² 

196 = 49r² 

r² = 4 

⇒ r = ±2

5. For r = 2 the terms are 14 – 6, 14 – 2, 14 + 2, 14 + 6 

⇒ 8, 12, 16, 20.

r = –2 gives the same four parts in reverse order.

The four parts are 8, 12, 16 and 20.