The sum of the 40 terms of an ap whose first term is 2 and common difference 4 will be?
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Answered by
4
Given,
Sn=40 terms
a = 2
d = 4
By the relation we have,
Sn = n/2(2a + (n-1)d)
40 = n/2 (2 x 2 + (n - 1) x4)
40 = n/2 (4 + 4n - 4)
40 =n/2 (4n)
40 = 4n2/2
80=4n2
n2 = 80/4
n2 =20
n =2 rootover 5
Sn=40 terms
a = 2
d = 4
By the relation we have,
Sn = n/2(2a + (n-1)d)
40 = n/2 (2 x 2 + (n - 1) x4)
40 = n/2 (4 + 4n - 4)
40 =n/2 (4n)
40 = 4n2/2
80=4n2
n2 = 80/4
n2 =20
n =2 rootover 5
Answered by
0
Answer:
Step-by-step explanation:
Answer is wrong
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