A little help required up there.
Number 12 please.
With explanation.
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Sn=3n²-2n,
then
S1=3(1)²-2×1=3-2=1=a,
S2=3(2)²-2×2=3×4-4=8,
then
d=8-1=7,
therefore
T19=a+(n-1)d,
=1+(19-1)×7,
=1+18×7,
=1+126=127
then
S1=3(1)²-2×1=3-2=1=a,
S2=3(2)²-2×2=3×4-4=8,
then
d=8-1=7,
therefore
T19=a+(n-1)d,
=1+(19-1)×7,
=1+18×7,
=1+126=127
allysia:
it's 3x^2 in the question
=_=
kk
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