Question

Proove that sum of later half of 2n terms of an ap if equal to one third the sum of first 3n terms

Answer 1

Sol:
Case 1
No. Of terms is 2n/n=2
First term of series (a)= (n+1) do term
Tn=a+(n-1)d
a+(n+1-1)d
=a+nd
Sum of later half 2n terms= n/2 [2a+(n-1) d]
=n/2 [2(a+nd)+(n-1)d]
=n/2 [2a+2nd+need]
=n/2 [2a+(3n-1)d]
Three times d sum of later half of 2n terms=3n/2 [2a+(3n-1]
sum of 3n terms of d same series=n/2
[2a+(n-1)d]
3n/2[2a+(3n-1)d]

Hence this is d sol for d question

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