+0

# help

0
47
3

The sum of four consecutive terms of an arithmetic progression is 32 and the ratio of product of the first and the last term to the product of two middle terms is 7:15.  Find the numbers. ​

Jan 31, 2020

#1
0

solve n + n + 1 + n + 2 + n + 3 = 32
n×(n + 3)/((n + 2) (n + 1)) = 7/15 for n

[NO SOLUTIONS EXIST !!]

Jan 31, 2020
#2
0

Mr. BB, you’re about as dumb as dumb can get.

How did you ever live so long?

Guest Jan 31, 2020
#3
+18250
0

Let the first term of the progression be a and let the common difference be d.

Then the terms of the progression are:  a,  a + d,  a + 2d,  a + 3d

The ratio is:  [a · (a + 3d)] / [(a + d) · (a + 2d)]  =  7/15

Cross-multiplying:   [a · (a + 3d)] ·15  =  7 · [(a + d) · (a + 2d)]

8a2 + 24ad - 14d2  =  0

4a2 + 12ad - 7d2  =  0

Solving for a using the quadratic formula with a = 4, b = 12d, and c = -7d2

a  =  [ -12d  +/-  sqrt{ (12d)2 - 4·(4)(-7d2) }  /  (2·4)

=  [ -12d  +/-  sqrt{ 144d2 + 112d2 } / 8

=  [ -12d  +/-  sqrt{ 256d2 } ] 8

=  [ -12d  +/- 16d ] / 8

I'm only going to consider the "+" answer:

=  [ -12d + 16d ] / 8

=  4d/8

=  d/2

Since the sum = 32:  a + (a + d) + (a + 2d) + (a + 3d)  =  32

(d/2) + (d/2 + d) + (d/2 + 2d) + (d/2 + 3d)  =  32

8d  =  32

d  =  4

Since  a  =  d/2     --->     a  =  4/2  =  2

With  a = 2  and  d = 4:       2, 6, 10, 14

I'll let you see what happens if you were to select the "-" answer.

Jan 31, 2020