Question

Find $n$ so that the $n^{th}$ terms of the following two A.P.'s are the same.
$1,7,13,19,....$ and $100,95,90,....$

Answer 1

The first arithmetic progression is $1,7,13,19,......$ where the first term is $a_1=1$, second term is $a_2=7$ and so on.

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

$d=a_2-a_1=7-1=6$

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$, therefore, the $n$th term of the first A.P is:

$T_n=1+(n-1)6=1+6n-6=6n-\mathrm{5.......}\left(1\right)$

Similarly, the second arithmetic progression is $100,95,90,,......$ where the first term is $a_1=100$, second term is $a_2=95$ and so on.

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

$d=a_2-a_1=95-100=-5$

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$, therefore, the $n$th term of the second A.P is:

$T_n=100+(n-1)(-5)=100-5n+5=-5n+\mathrm{105.......}\left(2\right)$

Now, since it is given that the $n$th terms of the two A.P's are equal therefore, equating equations 1 and 2 we get

$6n-5=-5n+105\Rightarrow 6n+5n=105+5\Rightarrow 11n=110\Rightarrow n=\frac{110}{11}\Rightarrow n=10$

Hence, $11$th term of the given sequence are equal.