five positive integer are in arithmetic progression. the sum of three middle number is 24 and product of first and fifth term is 48. find three terms of a.p

Question

how to find a.p when the quetion is technical

Answer 1

 

 

The first integer is N

The second Integer is N+k for some fixed number integer constant k

The third integer is N + 2*k = N + 2k

The fourth integer is N + 3*k = N + 3k

The fifth integer is N + 4*k = N + 4k

 

The sum of the 2nd, 3rd, and 4th, the three middle integers, is 24

N + k + N + 2k + N + 3k = 24

3*N + 6*k = 24 <--- combines like terms

N + 2k = 8  <---- divides everything by 3

N = 8 - 2k = 2(4-k) <--- solves for N; Please label this equation ALPHA

 

 

The product of the first and fifth is 48.

N(N+4k) = 48

 

N^2 + 4k*N = 48

 

N^2 + 4k*N - 48 = 0

 

(8 - 2k)^2 + 4k( 8 - 2k) - 48 = 0

 

64 - 32k + 4k^2 + 32k - 8k^2 - 48 = 0

 

-4k^2 + 16 = 0

 

-4( k^2 - 4) = 0

 

k^2 - 4 = 0

 

(k +2)(k-2) = 0

 

k+2 = 0 ---> k = -2    OR

k - 2 = 0 --->" k = 2

 

FOR k = -2, 

  N = 8 - 2k = 8 - 2(-2) = 8 + 4 = 12

     from equation ALPHA in bold above

 

The progression is 12, 10, 8, 6, 4.

The sum of the middle three is 10+8+6 = 24

and the product of the 1st and 5th is 12*4 = 48.

 

 

For k=2, N = 8 - 2*2 = 8 - 4 = 4.

The progression is 4,6,8,10,12.

The sum of the three middle 3 terms is 6+8+10 = 24

and the product of the 1st and 5th is 4*12 = 48.

 

 

 

So   4,6,8,10,12  and 12,10,8,6, 4