Given that,Sn=4n−n2First term, a=S1=4(1)−(1)2=4−1=3Sum of first two terms=S2=4(2)−(2)2=8−4=4Second term, a2=S2−S1=4−3=1Common difference (d)=a2−a=1−3=−2
The nthterm of an A.P is given by
an=a+(n−1)d=3+(n−1)(−2)=3−2n+2=5−2nTherefore, a3=5−2(3)=5−6=−1a10=5−2(10)=5−20=−15Hence, the sum of first two terms is 4.The second term is 1.The 3rd,10th, and nth terms are −1,−15, and 5−2n respectively.