The sum of four consecutive terms of an ap is the sum of the third and fourth terms is 11 find the terms
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Answer:
Given:-
Sum of 4 consecutive numbers of an AP
is 2.
Sum of the 3rd and 4th term is 11.
To Find:-
The terms
Solution:-
Define x:
Let the first term be a
First term = a
Second =a+d
Third term = a + 2d
Fourth term = a +3d
Form the equations:-
Sum of 4 consecutive numbers of an AP
is 2.
a +(a + d) + (a +2d) + (a + 3d) = 2
4a +6d =2
2a +3d =1-------------[1]
Sum of the 3rd and 4th term is 11
(a+2d)+ (a + 3d) = 11
2a +5d = 11---------------[2]
[2]-[1]:
2d = 10
d =5
Sub d=5 into [ 1]
2a +3(5)=1
2a +15 1
2a = -14
a = -7
Find the terms:-
First term =a
First ternm=-7
Second term = a +d
Second term= -7+5
Second term = -2
third term = a + 2d
third term =-7+2(5)
third term = 3
Fourth term = a +3d
Fourth term = -7 +3(5)
Fourth term = 8
Answer: The terms are -7, -2, 3 and 8.
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