Question

Find the largest positive term of the A.P. whose first two terms are $\frac{2}{5}$ and $\frac{12}{23}$.

Answer 1

Let the first term be $\frac{12}{13}$ and the second term be $\frac{2}{5}$.
Therefore, the common difference is $\frac{-14}{115}$.
Let the last term $T_n$ be the last term.
Therefore, $T_n=a+(n-1)d$
$\Rightarrow T_n=\frac{12}{23}+(n-1)\left(\frac{-14}{115}\right)$
$\Rightarrow T_n=\left(\frac{74-14n}{115}\right)$
Now, find $n$ such that it given the largest positive term.
For $n=1$, $\Rightarrow T_1=\frac{12}{23}$
For $n=2$, $\Rightarrow T_2=\frac{2}{5}$
For $n=3$, $\Rightarrow T_3=\frac{32}{115}$
For $n=4$, $\Rightarrow T_4=\frac{18}{115}$
For $n=5$, ${\Rightarrow T}_5=\frac{4}{115}$
For $n=6$, ${\Rightarrow T}_6=\frac{-2}{23}$
Notice that from $n=6$ onwards the terms in the series are negative.
Hence the largest positive term of the A.P. is $\frac{4}{115}$.
Thus, for $n=5$ the largest positive term obtained.