find the a.p whose third term is 16and 7th term exceeds the 5th term by 12
Answers
Step-by-step explanation:
Given:-
The third term is 16and 7th term exceeds the 5th term by 12 in an AP
To find:-
Find the AP?
Solution:-
Given that
The third term is 16and 7th term exceeds the 5th term by 12 in an AP
We know that
The first term is a and the common difference is d then the general term or nth term of the AP is
an = a+(n-1)d
Third term = a 3 = 16
=> a 3 = a+(3-1)d = 16
=> a 3 = a+2d = 16
a+2d = 16------------(1)
5 th term a 5 = a+(5-1)d
=> a 5 = a+4d
7th term = a 7 = a+(7-1)d
=> a 7 = a+6d
given that
7th term exceeds the 5th term by 12
=> a 7 = a 5 + 12
=> a +6d = a +4d +12
=> a+6d -a-4d = 12
=> (a-a)+(6d-4d) = 12
=> 0+2d = 12
=> 2d = 12
=> d = 12/2
=> d = 6
Common difference = 6
On Substituting the value of d in the equation (1)
=>a+2(6) = 16
=> a +12 = 16
=> a = 16-12
=> a = 4
First term of the AP = 4
Now,
The general form of an AP
a , a+d , a+2d , ....
a = 4
a+d = 4+6 = 10
a+2d = 4+2(6)=4+12=16
The AP : 4 , 10 ,16 ,...
Answer:-
The AP for the given problem is 4 , 10 ,16 ,...
Check:-
Third term = 16
5th term = 4+4(6)=4+24= 28
7th term = 4+6(6)=4+36=40
7th term - 5the term
=> 40-28
=> 12
Verified the given relations
Used formulae:-
1.The general form of an AP :a , a+d ,a+2d , ....
2.The first term is a and the common difference is d then the general term or nth term of the AP is
an = a+(n-1)d