Question

Find the sum to 16 terms of an AP where nth term is 3n-1

Answer 1

Dear student
$$\begin{gathered} \mathrm{Given},{\mathrm{n}}^{\mathrm{th}}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{sequence}{\mathrm{a}}_{\mathrm{n}}=3\mathrm{n}-1...\left(1\right) \\ \mathrm{Now},\mathrm{the}\mathrm{first},\mathrm{second},\mathrm{third}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{sequence}\mathrm{can}\mathrm{be}\mathrm{find}\mathrm{out}\mathrm{by}\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{n}=1,2,3\mathrm{respectively}\mathrm{in}\left(1\right). \\ \Rightarrow {\mathrm{a}}_1=3\left(1\right)-1=3-1=2 \\ \Rightarrow {\mathrm{a}}_2=3\left(2\right)-1=6-1=5 \\ \Rightarrow {\mathrm{a}}_3=3\left(3\right)-1=9-1=8 \\ \mathrm{Therefore},\mathrm{common}\mathrm{difference}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{sequence},\mathrm{d}={\mathrm{a}}_2-{\mathrm{a}}_1 \\ =5-2 \\ =3 \\ \mathrm{Now},\mathrm{sum}\mathrm{of}16\mathrm{terms}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{sequence}={\mathrm{S}}_{16} \\ \mathrm{and}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2{\mathrm{a}}_1+\left(\mathrm{n}-1\right)\mathrm{d}\right] \\ {\mathrm{S}}_{16}=\frac{16}{2}\left[2\left(2\right)+\left(16-1\right)3\right] \\ =8\left[4+\left(15\times 3\right)\right] \\ =8\left[4+45\right] \\ =8\times 49 \\ =392 \end{gathered}$$
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