how many terms of the AP 7, 14, 21... are needed to give the sum of 5740
Answers
Answered by
11
Answer:
Step-by-step explanation:
a= 7, d= 14-7 = 7, sn=5740
Sn= n/2(2a+(n-1)d
5740 x 2 = n( 2 x7 +(n-1)7)
11480 = n [ 14+7n -7) ]
11480= n (7n -7)
11480 =7n^2 -7n
7n^2 -7n - 11480 = 0
Dividing by 7
n^2 +n + 1640=0
By using quadratic formula
We get, n= 40
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Answered by
47
Answer:
40
Step-by-step explanation:
Here, a = 7
d = 14 - 7 = 7
Let the term of the AP needed to give the sum be n
We know that, sum of A.P (S) =
Here, S = 5740
Substituting the values, we get
→ 7n² + 7n - 11480 = 0
→7(n² + n - 1640) = 0
→ n² + n - 1640 = 0
→ n² + 41n - 40n - 1640 = 0
→ n(n + 41) - 40(n + 41) = 0
→(n + 41)(n - 40) = 0
Hence, n = - 41 or n = 40
Since, term can not be negative, n = 40
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