the sum of four consecutive terms of an arithmetic progression is 0.
The third term is 1 more than the sum of the first and the fourth term.
find the common difference d
Answers
Answer:
ANSWER
Let the 4 consecutive numbers in AP be
(a−3d),(a−d),(a+d),(a+3d)
According to the question,
a−3d+a−d+a+d+a+3d=32
4a=32
a=8
Now,
(a−d)(a+d)
(a−3d)(a+3d)
=
15
7
15(a
2
−9d
2
)=7(a
2
−d
2
)
15a
2
−135d
2
=7a
2
−7d
2
8a
2
=128d
2
Putting the value of a, we get,
d
2
=4
d=±2
So, the four consecutive numbers are 2,6,10,14 or 14,10,6,2.
Step-by-step explanation:
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Answer :
Common difference , d = 1
Note :
★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.
★ If a1 , a2 , a3 , . . . , an are in AP , then
a2 - a1 = a3 - a2 = a4 - a3 = . . .
Solution :
Let the four consecutive terms in AP be ;
a - 3d , a - d , a + d , a + 3d
Now ,
According to the question , the sum of these four consecutive terms is 0 .
Thus ,
=> (a - 3d) + (a - d) + (a + d) + (a + 3d) = 0
=> 4a = 0
=> a = 0/4
=> a = 0
Also ,
The 3rd term is 1 more than the sum of the 1st and 4th term .
Thus ,
=> (a + d) = (a - 3d) + (a + 3d) + 1
=> a + d = 2a + 1
=> 0 + d = 2×0 + 1
=> d = 1