If sum of n terms of an ap is 2n²+5.then prove that an =4n+3
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Let's take a general A.P. where the starting term is "a" and the common difference is "d".
Let S(n) denote the sum of N terms.
S(n) = (n(2a + (n-1)d))/2
= an + (n^2×d)/2 - (nd/2)
Group all alike terms
S(n) = n×(a-d/2) + n^2×(d/2)
Assuming you miswrote 2n^2 + 5n as just 5 which I think it should be, but I could be wrong...
d/2 = 2
d = 4
a - d/2 = 5
a = 5 + 2
a = 7
A(n) = a + (n-1)d
= 7 + (n-1)4
= 7 + 4n - 4
= 4n + 3
Hence proved.
Let S(n) denote the sum of N terms.
S(n) = (n(2a + (n-1)d))/2
= an + (n^2×d)/2 - (nd/2)
Group all alike terms
S(n) = n×(a-d/2) + n^2×(d/2)
Assuming you miswrote 2n^2 + 5n as just 5 which I think it should be, but I could be wrong...
d/2 = 2
d = 4
a - d/2 = 5
a = 5 + 2
a = 7
A(n) = a + (n-1)d
= 7 + (n-1)4
= 7 + 4n - 4
= 4n + 3
Hence proved.
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5
Brainly.in
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Secondary SchoolMath 5+3 pts
If sum of n terms of an ap is 2n²+5.then prove that an =4n+3
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Barthwalanoop8anoop · Ambitious
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Astrobolt
Astrobolt Expert
Let's take a general A.P. where the starting term is "a" and the common difference is "d".
Let S(n) denote the sum of N terms.
S(n) = (n(2a + (n-1)d))/2
= an + (n^2×d)/2 - (nd/2)
Group all alike terms
S(n) = n×(a-d/2) + n^2×(d/2)
Assuming you miswrote 2n^2 + 5n as just 5 which I think it should be, but I could be wrong...
d/2 = 2
d = 4
a - d/2 = 5
a = 5 + 2
a = 7
A(n) = a + (n-1)d
= 7 + (n-1)4
= 7 + 4n - 4
= 4n + 3
Hence proved.
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