The sum of 4 consecutive terms of an AP is 32 and the ratio of the product of the 1st and the last terms to the product of the 2 middle terms is 7:15.Find the number.
Answers
Common difference of AP = a and d
so a + (a+d) + (a+2d) + (a+3d) = 32
4a + 6d = 32
2a + 3d = 16 (smallest form)
2a = 16 - 3d
a = (16 -3d)/2 .............................(1)
the ratio of the product of the 1st and the last terms to the product of the 2 middle terms is 7:15.
∴ a(a+3d) = 7/5
(a+d)(a+2d)
15a (a+3d) = 7(a+d)(a+2d) (cross multipication)
15a² + 45ad = 7a² + 14ad + 7ad + 14d²
15a² + 45ad - 7a² - 14ad - 7ad - 14d²
= 8a² + 24ad - 14d²
= 4a² + 12ad - 7d²
= 4a² + 14ad - 2ad - 7d²
= 2a(2a + 7d) - d(2a + 7d)
= (2a-d) (2a+7d)
a = /2 and -7d/2 ...........(2)
subtitute the value from 1 to 2
a = d/2
d/2 = (16-3d)/2
d = 16 - 3d
4d = 16
d = 4
a= d/2 = 4/2 = 2
Thus number are 2, (2+4), (2+2x4), and (2+3x4)
= 2, 6. 10 & 14
and
a = ⁻7d/2
⁻7d/2 = (16-3d)/2
⁻7d = 16 - 3d
-4d = 16
d = ⁻4
a = -7d/2 = 28/2 = 14
Thus number are 14, (14+⁻4), (14+2x⁻4), and (14+3x⁻4)
= ⁻14, 10, 6, & 2