Question

Find four numbers in AP such that their sum is 24 and product 384

Answer 1

$\huge\bold\red{Answer}$

___________________

Ans:3 5 7 9

suppose the numbers are: (a-3d),(a-d),(a+d),(a+3d)

We know that sum=s=n/2{2(a-3d)+(n-1)*2d}

here,

s=24 and

n=4

so from this expression a=6

according to the question (a-3d)(a-d)(a+d)(a+3d)=24

solving this by putting a=6

we get d=1

putting the values of a and d,

we get numbers as follows 3,5,7,9

hope it help u

Answer 2

$\sf{\underline{Let\:the\:terms\:be:}}$

$\sf{(a - 3d), (a - d), (a + d) \:and\:(a + 3d)}$

$\sf{\underline{Given:}}$

Sum of the numbers in AP = 24

$\sf{\underline{So:}}$

$\sf{a - 3d + a - d + a + d + a + 3d = 24}$

$\sf{a + a + a + a - d - 3d + 3d = 24}$

$\sf{4a = 24}$

$\sf{a = 6}$

$\sf{\underline{And:}}$

The product of these four numbers is: 945

$\sf{\underline{Therefore:}}$

$\boxed{\sf{(a - 3d) (a - d) (a + d) (a + 3d) = 945}}$

$\sf{\underline{Putting\:the\:values\:of:}}$ a =6

$\sf{(6 - 3d) (6 - d) (6 + d) (6 + 3d) = 945}$

$\sf{(36 - {9d}^{2} )(36 - {d}^{2} ) = 945}$

$\sf{1296 - 360 {d}^{2} + 9 {d}^{2} = 945}$

$\sf{\underline{So:}}$

$\implies$ $\sf{9 {d}^{4} - 360 {d}^{2} + 1296 - 945 = 0}$

$\implies$ $\sf{9 {d}^{4} - 360 {d}^{2} + 351 = 0}$

$\sf{\underline{Note:}}$ Dividing it by 9 we get,

$\implies$ $\sf{{d}^{4} - 40 {d}^{2} + 39 = 0}$

$\implies$ $\sf{ {d}^{4} - 39 {d}^{2} - {d}^{2} + 39 = 0}$

$\implies$ $\sf{{d}^{2} ( {d}^{2} - 39) - 1( {d}^{2} - 39) = 0}$

$\implies$ $\sf{( {d}^{2} - 1)( {d}^{2} - 39) = 0}$

$\boxed{\sf{ {d}^{2} - 1 = 0}}$

$\implies$ $\sf{{d}^{2} = 1}$

$\implies$ $\sf{d = \sqrt{1}}$

$\implies$ $\sf{d = 1}$

$\boxed{\sf{{d}^{2} - 39 = 0}}$

$\implies$ $\sf{ {d}^{2} = 39}$

$\implies$ $\sf{d = \sqrt{39}}$

$\implies$ $\sf{d = 6.244}$

$\sf{\underline{Note:}}$ 6.244 is not possible, so d = 1

$\sf{\underline{So:}}$

$\boxed{\sf{a = 6\:and\:d = 1}}$

The required four numbers of the AP are:

(i) (6 - 3)

$\implies$ 6 - 3 = 3

(ii) (6 - 1)

$\implies$ 6 - 1 = 5

(iii) (6 + 1)

$\implies$ 6 + 1 = 7

(iv) (6 + 3)

$\implies$ 6 + 3 = 9

$\sf{\underline{Therefore:}}$

Four numbers in AP are 3, 5, 7 and 9.