[tex]\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=-5\\ n=10\\ a_{10}=13 \end{cases} \\\\\\ a_{10}=a_1+(10-1)d\implies 13=-5+(10-1)d \\\\\\ 13=-5+9d\implies 18=9d\implies \cfrac{18}{9}=d\implies 2=d[/tex]
Answer:
Common difference = 2
Step-by-step explanation:
[tex]n^{th}[/tex] term of an arithmetic progression is given by a + (n-1)d, where a is the first term and d is the common difference.
Here first term is given as -5
a = -5
Tenth term is 13
n = 10
[tex]n^{th}[/tex] term = 13
We have
-5 + (10-1)d = 13
9 d = 13 + 5
9 d = 18
d = 2
Common difference = 2
