Question

the sum of four consecutive number is an ap is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15 find the numbers

Answer 1

Let (a - 3d), (a-d),(a+d),(a+3d) are

four consecutive terms in A.P.

Sum of 4 terms = 32 [ given ]

=> a-3d+a-d+a+d+a+3d = 32

=> 4a = 32

=> a = 32/4

=> a = 8 ----( 1 )

Product of first and last term : product

of middle terms = 7 : 15

=> (a-3d)(a+3d) : (a-d)(a+d) = 7: 15

=> (a² - 9d² ) : ( a² - d² ) = 7 : 15

=> 15(a² - 9d² ) = 7( a² - d² )

=> 15a² - 135d² = 7a² - 7d²

=> 15a² - 7a² = 135d² - 7d²

=> 8a² = 128d²

=> ( 8 × 8² ) = 128d²

=> ( 8 × 8 × 8 ) /128 = d²

=> 4 = d²

Therefore ,

d = ± √4

d = ± 2 ,

Now ,

Required terms are ,

i ) if a = 4 , d = 2 then

a - 3d = 4 - 3×2 = -2

a - d = 4 - 2 = 2 ,

a + d = 4 + 2 = 6 ,

a + 3d = 4 + 3 × 2 = 10,

ii ) if a = 4 , d = -2 ,

the four terms are ,

10, 6 , 2 , -2

••••

Answer 2

Answer:

Step-by-step explanation:

Suppose the numbers are x, x+d, x+2d, x+3d in the AP series.

now the sum of 4 numbers is 4x+6d = 32.

according to the question, x(x+3d)/(x+d)(x+2d)= 7/15

or, 15x^2 + 45dx = 7x^2 + 21dx +14d^2

or, 8x^2 +24dx -14d^2 = 0

or, 4x^2 + 12dx -7d^2= 0

or, (2x-d) (2x+7d) = 0

we will take the +ve value of x i.e d/2.

now, put the value of x in equation 4x+6d = 32 and will get d= 4 and so, x= 2

so the numbers are 2, 2+4 = 6, 2+8 = 10, and 2+12 = 14