Question

sum of 4 consecutive numbers in a AP is 32 and ratio of product if first and last terms to product of 2 middle terms is 7:15

Answer 1

Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d)
So, according to the question.a-3d + a - d + a + d + a + 3d =
324a = 32a = 32/4a = 8 ......(1)
Now, (a - 3d)(a + 3d)/(a - d)(a + d)
= 7/1515(a² - 9d²) =
7(a² - d²)15a² - 135d² =
7a² - 7d²15a² - 7a² =
135d² - 7d² 8a² =
128d²Putting the value of a = 8 in above we get.
8(8)² = 128d²
128d² = 512d² =
512/128d² = 4d = 2
So, the four consecutive numbers are 8 - (3*2)8 - 6 = 28 - 2 = 68 + 2 = 108 + (3*2)8 + 6 = 14
Four consecutive numbers are 2, 6, 10 and 14

Answer 2

Let the four consecutive numbers in AP be a-3d,a-d,a+d,a+3d.

then, a-3d+a-d+a+d+a+3d=32
        4a=32
        a=8 1
        (a-3d)(a=3d) / (a-d)(a+d)=7/15
        15(a2-9d2) = 7(a2-d2)
         15a2-135d2 = 7a2-7d2
          8a2-128d2 = 0
          d2 = 4, ±2
Therefore d=4  or d=±2 1mark
So, when a=8 and d=2, the numbers are 2,6,10,14.
When a=8,d=-2 the numbers are 14,10,6,2