The sum of four consecutive numbers in an A.P is 32 and the ratio of the product of the first and last and term to the product of two middle terms is 7:15. Find the numbers.
Answers
Answer:
The four consecutive terms of the AP are 2, 6, 10, 14
Step-by-step explanation:
Given, the sum of four consecutive numbers in an AP is 32.
Ratio of the product of the first and the last terms to the product of the two middle terms is 7:15.
We have to find the number.
Let the 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 = 32/4
a = 8
Product of the first and last terms = (a - 3d)(a + 3d)
Put a = 8,
= (8 - 3d)(8 + 3d)
= 64 - 9d²
Product of the two middle terms = (a - d)(a + d) = (a² - d²)
Put a = 8,
= 64 - d²
(64 - 9d²)/(64 - d²) = 7/15
15(64 - 9d²) = 7(64 - d²)
15(64) - 135d² = 7(64) - 7d²
15(64) - 7(64) = 135d² - 7d²
64(15 - 7) = (135 - 7)d²
64(8) = 128d²
d² = 64(8)/128
d² = 8/2
d² = 4
Taking square root,
d = ±2
Take d = 2,
When a = 8 and d = 2,
The four consecutive numbers are
a - 3d = 8 - 3(2) = 8 - 6 = 2
a - d = 8 - 2 = 6
a + d = 8 + 2 = 10
a + 3d = 8 + 3(2) = 8 + 6 = 14
Therefore, the four consecutive terms of the AP are 2, 6, 10, 14.