sum of first n terms of ap is an(n-1).Sum of squares of these terms is?
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sum of n terms of an AP = n/2 { 2A + (n-1)D} = An + n² D/2 - nD/2
= Dn²/2 + n( A - D/2) comparing this with one given in ques i.e. an² - an
⇒ D/2 = a ⇒ D = 2a and A - D/2 = - a ⇒ A = 0
so first term of this AP is 0
and nth term = 0 +(n-1)D = (n-1)2a = 2an - 2a
Tn = 2an - 2a (Tn) ² = 4a²n² + 4a² - 8a²n
sum of (Tn)² = ∑(Tn)² = ∑ 4a² n² + 4a² -8a²n
= 4a²∑n² + 4a²∑1 -8a²∑n
= 4a² n(n+1)(2n+1)/6 + 4a²n - 8a²n(n+1)/2
= 2a²n(n+1)(2n+1)/3 + 4a²n - 4a²n(n+1)
as sum of squares of n (natural nos.) = n(n+1)(2n+1)/6
and that of sum of n (natural nos.) = n(n+1)/2
also sum 1 n times = n
hope my ans is correct .............if any query, then ask.
= Dn²/2 + n( A - D/2) comparing this with one given in ques i.e. an² - an
⇒ D/2 = a ⇒ D = 2a and A - D/2 = - a ⇒ A = 0
so first term of this AP is 0
and nth term = 0 +(n-1)D = (n-1)2a = 2an - 2a
Tn = 2an - 2a (Tn) ² = 4a²n² + 4a² - 8a²n
sum of (Tn)² = ∑(Tn)² = ∑ 4a² n² + 4a² -8a²n
= 4a²∑n² + 4a²∑1 -8a²∑n
= 4a² n(n+1)(2n+1)/6 + 4a²n - 8a²n(n+1)/2
= 2a²n(n+1)(2n+1)/3 + 4a²n - 4a²n(n+1)
as sum of squares of n (natural nos.) = n(n+1)(2n+1)/6
and that of sum of n (natural nos.) = n(n+1)/2
also sum 1 n times = n
hope my ans is correct .............if any query, then ask.
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