Question

the sum of four terms of an A.P.Is 24 and their product is 945.find the terms

Answer 1

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Here's ur answer :-


Let the terms be (a - 3d), (a - d), (a + d) and (a + 3d)
Given : Sum of the numbers in AP = 24
So, 
a - 3d + a - d + a + d + a + 3d = 24
4a = 24
a = 6
And the product of these four numbers is 945
So,
(a - 3d) (a - d) (a + d) (a + 3d) = 945
Putting the value of a =6
(6 - 3d) (6 - d) (6 + d) (6 + 3d) = 945
(36 - 9d²) (36 - d²) = 945
1296 - 360d² + 9d⁴ = 945
9d⁴ - 360d² + 1296 - 945 = 0
9d⁴ - 360d² + 351 = 0   Dividing it by 9 we get
d⁴ - 40d² + 39 = 0
d⁴ - 39d² - d² + 39 = 0
d²(d² - 39) - 1(d² - 39) = 0
(d² - 1) (d² - 39) = 0
d² - 1 = 0
d² = 1
d = √1
d = 1
d² - 39 = 0
d² = 39
d = √39
d = 6.244  
d = 6.244 is not possible, so d = 1
So, a = 6 and d = 1
The required four numbers of the AP are (6 - 3), (6 - 1), (6 + 1) and (6 + 3)
= 3, 5, 7 and 9


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Answer 2

Let the Four terms of AP be
a-3d,a-d,a+d,a+3d
Sum of these numbers
a-3d+a-d+a+d+a+3d=24
4a=24
a=6

[tex]product \\ ( {a}^{2} - 9 {d}^{2} )( {a}^{2} - {d}^{2} ) = 945 \\ (36 - 9 {d}^{2} )(36 - {d}^{2} ) = 945 \\ 1296 - 36 {d}^{2} - 324 {d}^{2} + 9 {d}^{4} = 945 \\ 351 = -361{d}^{2}+9{d}^{4}//

hence numbers are
3,5,7,9
or 9,7,5,3