For each positive integer n, the nth positive triangular number is equ

\({T_n} = 1 + 2 + \ldots + n\,\,\,\,\left( {n \ge 1\,\,{\mathop{\rm int}} } \right)\)

\({T_n} = {{n \cdot \left( {n + 1} \right)} \over 2}\,\,\,\,\,\,\left[ {{\rm{arithmetic}}\,\,{\rm{sequence}}} \right]\)

\(?\,\,\,:\,\,\,\underline {{\rm{not}}} \,\,\,T\)


\(\left( A \right)\,\,\left\{ \matrix{\\
\,n \cdot \left( {n + 1} \right) = 2 \cdot 105 = 210 \hfill \cr \\
\,15 \cdot 15 = 225 \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{try}}!} \,\,\,\,{{14 \cdot 15} \over 2} = 7 \cdot 15 = 105\,\,\,{\rm{works}}!\,\,\,\,\, \Rightarrow \,\,\,\,\,105 = {T_{14}}\)

\(\left( B \right)\,\,\left\{ \matrix{\\
\,n \cdot \left( {n + 1} \right) = 2 \cdot 210 = 420 \hfill \cr \\
\,20 \cdot 20 = 400 \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{try}}!} \,\,\,\,{{20 \cdot 21} \over 2} = 10 \cdot 21 = 210\,\,\,{\rm{works}}!\,\,\,\,\, \Rightarrow \,\,\,\,\,210 = {T_{20}}\)

\(\left( C \right)\,\,\left\{ \matrix{\\
n \cdot \left( {n + 1} \right) = 2 \cdot 300 = 600 \hfill \cr \\
25 \cdot 25 = 625 \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{try}}!} \,\,\,\,{{24 \cdot 25} \over 2} = 12 \cdot 25 = 300\,\,\,{\rm{works}}!\,\,\,\,\, \Rightarrow \,\,\,\,\,300 = {T_{24}}\)

\(\left( D \right)\,\,\,{T_{24}} < 311 < 300 + 25 = {T_{24}} + 25 = {T_{25}}\)


The correct answer is (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.