the sum of 4 consequetive term which are in a ap is 32 and the product of the first and last term to the product of two middle terms is 7:15 find the number
Answers
Answer:
The terms are 2, 6, 10, 14
Step-by-step explanation:
Since consecutive terms are in A.P, then let the terms be
a, (a+d), (a+2d), (a+3d) where d is the difference of series.
From question,
a+(a+d)+(a+2d)+(a+3d) = 32
4a + 6d = 32
2(2a + 3d) = 32
2a + 3d = 16
a = (16-3d)/2 -----------------(i)
Again from question, 1st X 4th : 2nd X 3rd = 7/15
{a(a+3d)} / {(a+d)(a+2d)} = 7/15
15a(a+3d) = 7(a+d)(a+2d)
15a² + 45ad = 7a²+21ad+14d²
15a²-7a²+45ad-21ad-14d² = 0
8a²+24ad-14d² = 0
2(4a²+12ad-7d²) = 0
4a²+14ad-2ad-7d² = 0
2a(2a+7d) -d(2a+7d) = 0
(2a-d) (2a+7d) = 0
2a-d = 0 , 2a+7d = 0
a = d/2 , a = -7d/2
let's take a=d/2 and put in eq (i),
d/2 = (16-3d)/2
2d = 32-6d
2d+6d = 32
d = 32/8 = 4
Since a = d/2,
a = 4/2 = 2
a = 2, d = 4
put this value in the above series, then it becomes
2, (2+4), (2+2X4), (2+3X4)
2, 6, 10, 14