The value of x for which 3x,(x+8) and (5x+2) are three
consecutive terms of an A.P. is
Answers
Step-by-step explanation:
Given:-
3x,(x+8) and (5x+2) are three consecutive terms of an A.P.
To find:-
Find the value of x ?
Solution:-
Method-1:-
Given that
3x,(x+8) and (5x+2) are three consecutive terms of an A.P.
Since they are in the AP then the common difference is same.
=>Common difference = tn - t(n-1)
=> (x+8)-(3x) = (5x+2)-(x+8)
=>x+8-3x = 5x+2-x-8
=> 8-2x = 4x-6
=> 8+6 = 4x+2x
=> 14 = 6x
=> 6x = 14
=> x = 14/6
=> x = 7/3
Therefore,x=7/3
Method-2:-
Given that
3x,(x+8) and (5x+2) are three consecutive terms of an A.P.
We know that
a,b,c are the three consecutive terms in an AP then 2b = a+c
we have
a = 3x
b=x+8
c=5x+2
=> 2(x+8) = 3x+5x+2
=> 2x+16 = 8x+2
=> 16-2 = 8x-2x
=> 14 = 6x
=> 6x = 14
=> x = 14/6
=> x = 7/3
Therefore,x=7/3
Answer:-
The value of x for the given problem is 7/3
Used formulae:-
1.Common difference in an AP = tn - t(n-1)
2.a,b,c are the three consecutive terms in an AP then 2b = a+c or b = (a+c)/2