If for a sequence (tn),sn=2n²+5n,find tn and show that the sequence is an a.p

Answer 1

s(1) = t(1) = 2+5 = 7
s(2) = 2(2^2) + 5(2) = 18
s(3) = 2(9) + 5(3) = 33
s(4) = 2(16) + 5(4) = 52
..

t(2) = s(2) - s(1) = 18-7 = 11
t(3) = s(3) - s(2) = 33-18
= 15
t(4) = s(4) - s(3) = 52-33 = 19

so we have:
7,11,15,19,... looks like an AP with
a = 7 , d = 4
term(n) = a + (n-1)d
= 7 + 4(n-1)
= 7 + 4n - 4
= 4n + 3

in general:
term(n) = sum(n) - sum(n-1)
= 2n^2 + 5n - (2(n-1)^2 + 5(n-1) )
= 2n^2 + 5n - ( 2n^2 - 4n + 2 + 5n - 5)
= 2n^2 + 5n - 2n^2 -n + 3
= 4n + 3