Find 4 consecutive integers such that twice the sum of the two largest integers is 91 more than 3 times the first integer.

Question

Find 4 consecutive integers such that twice the sum of the two largest integers is 91 more than 3 times the first integer.

Answer 1

Here's what we know:

b = a + 1

c = a + 2

d = a + 3

 

2(c + d) = 3a + 91

 

We have values for c and d so plug them in:

2(a + 2 + a + 3) = 3a + 91

2(2a + 5) = 3a + 91

4a + 10 = 3a + 91

Subtract 3a:

a + 10 = 91

subtract 10:

a = 81

 

so we have 81, 82, 83, 84

 

check:

2(83 + 84) = 3 * 81 + 91

2 * 167 = 243 + 91

334 = 334