plz help in this question
Answers
Answer:
8,12,16 & 20
Step-by-step explanation:
Let take first term of AP as 'a'
and let d be the difference
First term = a
2nd term = a+d
3rd term= a+2d
4th term= a+3d
We have given that ratio of product of 1st and 4th and 2nd and 3rd term is 5/6
i.e.
a×(a+3d)/(a+d)×(a+2d)= 5/6
6a(a+3d)= 5(a+d)×(a+2d)
6a²+18ad= 5a²+15ad+10d²
a²+3ad-10d²=0
a²+5ad-2ad-10d²=0
a(a+5d)-2d(a+5d)= 0
(a+5d)(a-2d)=0
a=2d ........................(1)
Further,
sum of all terms is 56
a+a+d+a+2d+a+3d= 56
4a+6d= 56 .........(2)
putting the value of (1) in (2)
4a+ 3a= 56
a= 8
first term=8
2nd term= 8+4=12
3rd term=8+4×2= 16
4th term= 8+4×3= 20
Answer:
The 4 numbers are 8 , 12 , 16 , 20
or 20 , 16 , 12 , 8
Step-by-step explanation:
As, 4 numbers are in AP so, we can take them as -
a-3d , a-d , a+d , a+3d
It is given that sum of these 4 numbers is 56 so -
(a-3d)+(a-d)+(a+d)+(a+3d) = 56
=> 4a = 56
=> a = 14
Also, it is given that ratio of product of extremes to product of means is 5 :6
=> [(a-3d) * (a+3d)]/[(a-d) * (a+d)] = 5/6
=> [(14-3d) * (14+3d)]/[(14-d) * (14+d)] = 5/6
=> (196 - 9d²)/(196 - d²) = 5/6
=> (1176 - 54d²) = (980 - 5d²)
=> 196 = 49d²
=> d² = 4
=> d = 2 or -2
So, the 4 numbers are = a-3d , a-d , a+d , a+3d
= 14-3*2 , 14-2 , 14+2 , 14+3*2
= 8 , 12 , 16 , 20
Also, the 4 numbers can be = a-3d , a-d , a+d , a+3d
= 14-3*(-2) , 14+2 , 14-2 , 14+3*(-2)
= 20 , 16 , 12 , 8