Question

Find four number in A.P such that the sum of the first and last number is 8 and the product of second And third number is 12 give solution

Answer 1

let the four terms be
a-2d,a-d,a+d,a+2d
now, it's given that a+2d +a-2d=8
2a=8
a=4
also we are given that,
(a-d)(a+d)=a^2-b^2=12
(4)^2-d^2=12
16-12=d^2
4=d^2
d=2
now the four no. are 0,2,6,8

Answer 2

Answer:

Numbers are -2 , 2 , 6 , 10.

Step-by-step explanation:

Given: Sum of 1st no. & last no. = 8

          Product of 2nd & 3rd no. = 12

To find: The nos.

Let the 4 nos of AP are a - 3d , a - d , a + d , a + 3d

According to question,

( a - 3d ) + ( a + 3d ) = 8

2a = 8

$a=\frac{8}{2}$

a = 4

( a - d ) × ( a + d ) = 12

a² - d² = 12

4² - d² = 12

16 - d² = 12

d² = 16 - 12

d² = 4

d = ±√4

d = ± 2

So, nos. are

when d = +2

a - 3d = 4 - 3 × 2 = 4 - 6 = -2

a - d = 4 - 2 = 2

a + d = 4 + 2 = 6

a + 3d = 4 + 3 × 2 = 4 + 6 = 10

when d = -2

a - 3d = 4 - 3 × (-2) = 4 + 6 = 10

a - d = 4 - (-2) = 6

a + d = 4 + (-2) = 2

a + 3d = 4 + 3 × (-2) = 4 - 6 = -2

Therefore, Numbers are -2 , 2 , 6 , 10.