Question

Find the term of the series $25,$ $22\frac{3}{4},20\frac{1}{2},18\frac{1}{4},...$ which is numerically the smallest.

Answer 1

$25,22\frac{3}{4},20\frac{1}{2},18\frac{1}{4},......$

common difference $d=22\frac{3}{4}-20\frac{1}{2}=\frac{9}{4}{T}_1=25,T_2=22\frac{3}{4},T_3=20\frac{1}{2},T_4=18\frac{1}{4}{T}_5=18\frac{1}{4}-\frac{9}{4}=\frac{73}{4},T_6=-\frac{73}{4}-\frac{9}{4}=\frac{64}{4},T_7=\frac{64}{4}-\frac{9}{4}=\frac{55}{4},T_8=\frac{55}{4}-\frac{9}{4}=\frac{46}{4},T_9=\frac{46}{4}-\frac{9}{4}=\frac{37}{4},T_{10}=\frac{37}{4}-\frac{9}{4}=\frac{28}{4},T_{11}=\frac{28}{4}-\frac{9}{4}=\frac{19}{4},T_{12}=\frac{19}{4}-\frac{9}{4}=\frac{10}{4}$

$12$ is the number in the series which is numerically the smallest