# 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.

• 3

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

So,
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
• 251

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

So,
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

• 31

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

So,

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

• 79

How do u get d^2

• 5

Thank you for the answer. Thumps up for you.

• 21
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