Question
What is the $$15^{th}$$ term of the series $$3, 7, 13, 21, 31, 43, . . . . . \,?$$
A.
205
B.
225
C.
238
D.
241
Answer :
241
Solution :
Let,
\[\frac{\begin{gathered}
S = 3 + 7 + 13 + 21 + 31 + \,\,\,.....\,\,\, + {a_n} \hfill \\
\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,- S = \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,3 + 7 + 13 + 21 + 31 + .... + {a_{n - 1}} + {a_n} \hfill \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, - \,\,\,\,\, - \,\,\,\,\, - \,\,\,\,\,\, - \,\,\,\,\,\,\, - \,\,\,\,\,\, - \,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\, - \hfill \\
\end{gathered} }{{0 = 3 + 4 + 6 + 8 + 10 + 12 + ..... - {a_n}\,\,\,\,\,\,\,\,\,\,\,}}\]
$$\eqalign{
& \Rightarrow {a_n} = 3 + \left[ {4 + 6 + 8 + 10 + 12 + .....\,\left( {n - 1} \right){\text{terms}}} \right] \cr
& = 3 + \frac{{\left( {n - 1} \right)}}{2}\left[ {8 + \left\{ {\left( {n - 1} \right) - 1} \right\} \times 2} \right] \cr
& = 3 + \frac{{\left( {n - 1} \right)}}{2}\left[ {8 + 2n - 4} \right] \cr
& = 3 + \frac{{\left( {n - 1} \right)}}{2}\left( {2n + 4} \right) \cr
& = 3 + \left( {n - 1} \right)\left( {n + 2} \right) \cr
& \therefore {15^{th}}{\text{ term }} = {a_{15}} = 3 + \left( {15 - 1} \right)\left( {15 + 2} \right) \cr
& = 3 + 14 \times 17 = 241 \cr} $$