find the ap of which sum of 2nd term and 3 term is 22 and product of 1st and 4th term is 85
Answers
Step-by-step explanation:
Given :-
In an AP, Sum of 2nd and 3rd terms is 22.
Product of 1st and 4th terms is 85.
To find:-
Find the AP ?
Solution:-
Let the first term of an AP = a
Let the common difference = d
The general form of the AP = a , a+d , a+ 2d ,...
We know that
nth term of the AP = an = a+(n-1)d
Now
2nd term = a2 = a+d
3rd term = a3 = a+2d
4th term =a4 = a+3d
Sum of 2nd and 3rd terms = 22
=> a2 + a3 = 22
=> a+d +a+2d = 22
=> 2a +3d = 22
=> 3d = 22-2a-----------(1)
=> d = (22-2a)/3-----------(2)
And Product of 1st and 4th terms = 85
=> a(a+3d) = 85
=> a(a+22-2a) = 85 (from (1))
=> a(22-a) = 85
=> 22a -a^2 = 85
=> 22a - a^2 -85 = 0
=> -(22a+a^2+85) = 0
=> a^2 - 22a + 85 = 0
=> a^2 - 17 a - 5a + 85 = 0
=> a(a-17)-5(a-17) = 0
=> (a-17)(a-5) = 0
=> a -17 = 0 or a - 5 = 0
=>a = 17 or a = 5
If a = 17 then (2) becomes
d = (22-2(17))/3
=> d = (22-34)/3
=> d = -12/3
=> d = -4
If a = 5 then (2) becomes
d = (22-2(5))/3
=> d = (22-10)/3
=> d = 12/3
=> d = 4
If a = 17 and d = -4 then AP : 17, 17-4, 17+2(-4)
=> 17 , 13 , 9
If a = 5 and d = 4 then AP : 5, 5+4 , 5+2(4)
=> 5, 9, 13
Answer:-
The AP is 5,9,13,17... or 17, 13,9,5,...
Used formulae:-
- The general form of the AP = a , a+d , a+ 2d ,...
- nth term of the AP = an = a+(n-1)d
- a = First term
- d = Common difference
- n=Number of terms