Question

Which term of the arithmetic sequence $24,23\frac{1}{4},22\frac{1}{2},21\frac{3}{4},....$ is $3$?

Answer 1

The given arithmetic progression is $24,23\frac{1}{4},22\frac{1}{2},21\frac{3}{4},......$ where the first term is $a_1=24$, second term is $a_2=23\frac{1}{4}$ and so on.

We find the common difference $d$ by subtracting the first term from the second term as shown below:

$d=a_2-a_1=23\frac{1}{4}-24=\frac{93}{4}-24=\frac{93-96}{4}=-\frac{3}{4}$

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$

Let the $n$th term of the given A.P be $T_n=3$ and substitute $a=24$ and $d=-\frac{3}{4}$ in $T_n=a+(n-1)d$ as follows:

$T_n=a+(n-1)d\Rightarrow 3=24+(n-1)(-\frac{3}{4})\Rightarrow 3-24=(n-1)(-\frac{3}{4})\Rightarrow -21=-\frac{3}{4}n+\frac{3}{4}$

$\Rightarrow \frac{3}{4}n=\frac{3}{4}+21\Rightarrow \frac{3}{4}n=\frac{3+84}{4}\Rightarrow \frac{3}{4}n=\frac{87}{4}\Rightarrow n=\frac{87}{4}\times \frac{4}{3}\Rightarrow n=29$

Hence, $29$th term of the given A.P is $3$.