Question

The sum of four consecutive number in an AP is 32 and the ratio is the product of the first and last term to the product is two middle terms is 7:15 . Find the number?

Answer 1

Smash❕ bash❕


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 = 32
4a = 32
a = 32/4
a = 8 ......(1)
Now, (a - 3d)(a + 3d)/(a - d)(a + d) = 7/15
15(a² - 9d²) = 7(a² - d²)
15a² - 135d² = 7a² - 7d²
15a² - 7a² = 135d² - 7d²
8a² = 128d²
Putting the value of a = 8 in above we get.
8(8)² = 128d²
128d² = 512
d² = 512/128
d² = 4
d = 2
So, the four consecutive numbers are
8 - (3*2)
8 - 6 = 2
8 - 2 = 6
8 + 2 = 10
8 + (3*2)
8 + 6 = 14
Four consecutive numbers are 2, 6, 10 and 14



. Hope it helps

Answer 2

Hi!
Here's your answer>>>>>>
S4=32
n/2(2a+(n-1)d)=32
4/2(2a+(4-1)d)=32
2(2a+3d)=32
2a+3d=16
a=16-3d/2 (1)

a(a+3d)/a+d(a+2d)=7/15
a²+3ad/a²+2ad+ad+2d²=7/15
15(a²+3ad)=7(a²+3ad+2d²)
15a²+45ad=7a²+21ad+14d²
15a²-7a²+45ad-21ad-14d²=0
8a²+24ad-14d²=0
Put Equation 1 in this Equation
8(16-3d/2)²+24×16-3d/2-14d²=0
8×(16-3d)²/4+12(16-3d)-14d²=0
2(256+9d²-96d)+192-36d-14d²=0
512+18d²-192d+192-36d-14d²=0
18d²-14d²-192d-36d+512+192=0
4d²-228d+704=0
4(d²-57d+176)=0
d²-57d+176=0
Then do Factorisation using Middle Term splitting
You will get d than put it in Equation 1 and u will get a