Math, asked by 9910148152, 1 year ago

If sum of n terms of an ap is 2n²+5.then prove that an =4n+3

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Answered by Astrobolt
10
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.
Answered by barthwalanoop8anoop
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

Report by 9910148152 17.02.2018

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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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