Question

Sn= 2n²+3n then d
please be solve ​

Answer 1

sn =2n2+3n

TT=sn−sn−1

=2n2+3n−2(n−1) 2−3(n−1)

=2n2 +3n−2n 2+2n−2−3n+3

=3n−2+n+3

=4n+1

T1 =4+1=5

TT =8+1=9

d=−T1+T 2

d=9−5=4

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Answer 2

Answer:

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