Question

How many terms are there in the following Arithmetic Progression?
(i) $-1,-\frac{5}{6},-\frac{2}{3},...,\frac{10}{3}$
(ii) $7,13,19,....,205$

Answer 1

(i) The given arithmetic progression is $-1,-\frac{5}{6},-\frac{2}{3},........,\frac{10}{3}$ where the first term is $a_1=-1$, second term is $a_2=-\frac{5}{6}$ and the $n$th term is $T_n=\frac{10}{3}$.

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

$d=a_2-a_1=-\frac{5}{6}-(-1)=-\frac{5}{6}+1=\frac{-5+6}{6}=\frac{1}{6}$

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

To find the number of terms of the A.P, substitute $a=-1$, $T_n=\frac{10}{3}$ and $d=\frac{1}{6}$ in $T_n=a+(n-1)d$ as follows:

$T_n=a+(n-1)d\Rightarrow \frac{10}{3}=-1+(n-1)\left(\frac{1}{6}\right)\Rightarrow \frac{10}{3}+1=(n-1)\left(\frac{1}{6}\right)\Rightarrow \frac{10+3}{3}=\frac{1}{6}n-\frac{1}{6}\Rightarrow \frac{1}{6}n=\frac{13}{3}+\frac{1}{6}$

$\Rightarrow \frac{1}{6}n=\frac{13\times 2}{3\times 2}+\frac{1}{6}\Rightarrow \frac{1}{6}n=\frac{26}{6}+\frac{1}{6}\Rightarrow \frac{1}{6}n=\frac{27}{6}\Rightarrow n=\frac{27}{6}\times \frac{6}{1}\Rightarrow n=27$

Hence, there are $27$ terms in the given arithmetic progression.

(ii) The given arithmetic progression is $7,13,19,.....,205$ where the first term is $a_1=7$, second term is $a_2=13$ and the $n$th term is $T_n=205$.

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

$d=a_2-a_1=13-7=6$

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

To find the number of terms of the A.P, substitute $a=7$, $T_n=205$ and $d=6$ in $T_n=a+(n-1)d$ as follows:

$T_n=a+(n-1)d\Rightarrow 205=7+(n-1)6\Rightarrow 205=7+6n-6\Rightarrow 205=6n+1\Rightarrow 205-1=6n\Rightarrow 6n=204\Rightarrow n=\frac{204}{6}\Rightarrow n=34$

Hence, there are $34$ terms in the given arithmetic progression.