Question

Q2.

The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7:15. Find the numbers​

Answer 1

Answer:

the four terms are ,

10, 6 , 2 , -2

Step-by-step explanation:

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