Mathematics - Sequence & Series Question with Solution | TestHub

The first term $a_1=S_1=3(1{)}^2+2(1)=5$. So (P)-(1).$S_2=3(2{)}^2+2(2)=16$. The second term $a_2=S_2-S_1=16-5=11$. The common difference $d=a_2-a_1=11-5=6$. So (Q)-(2). The $n$ -th term $a_n=S_n-S_{n-1}=(3n^2+2n)-(3(n-1{)}^2+2(n-1))=6n-1$. The 5th term $a_5=6(5)-1=29$. So (R)-(3). The sum of the first 10 terms $S_{10}=3(10{)}^2+2(10)=300+20=320$. So (S)-(4).