Question

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

Answer 1

$\huge \underline \bold \color{plum} \mathcal{HEYA }$

Your answer along with full explanation is in the pic....

Hope it helps...

Answer 2

Answer:

Step-by-step explanation:

HI,

The sum of 4 consecutive number in an A.P is 32

So, the 4 consecutive numbers are a,(a+d),(a+2d),(a+3d)

Now, a + a+d + a+2d + a+3d = 32

= 4a + 6d = 32

= 2a + 3d = 16 { Half the number }  

Now, we need to find the product of the 1st term and the last term,

So,    ( a-3d/2) (a-d/2) (a+d/2) (a+3d/2)

Now, 4a = 32

  Therefore, a = 32/4

 

          a = 8

Now, Substituting the value of 8 in 'a'

         

→ ( 8-3d/2) (8 - d/2) (8 + d/2) ( 8+3d/2) -------- Equation (1)

↔TWO MIDDLE TERM IS 7:15

Now, ( 8 - 3d/2 ) ( 8 + 3d/2) / (8 - d/2) (8 + d/2) = 7/15  

 ⇔ Now, cross-multiply, and apply the formula a² - b² = (a - b) (a + b)

→ 15 ( 64 - 9d²/4 ) = 7 ( 64 - d²/4) {∴ 15 × 64 = 960 and 15 × 9 = 135 }

→ [ 960 - 135 d²/4 ]= 448 - 7d²/4

→ 960 - 448 = 135d²/4 - 7d²/4 {∵ 135 - 7 = 128 } and L.C.M will be 4

→ 512 = 128/4 d²

Now, d² = 16

d = 4

Now, substitute the value of d = 4 in eq.1

( 8 - 3d/2 ) (8 - d/2) (8 + d/2) ( 8+3d/2)  

The numbers are = 2 , 6 , 10 , 14

Hope it helps you !

PLZ THANK ME AND FOLLOW