In Arithmetic sequence
a13=4;a21=8;a5=?
What will be the answer?? help
Lets assume this question is: We have an arithmetic progression $\displaystyle \{a_i,\ i=1,2,..\}$, and $\displaystyle a_{13}=4$ and $\displaystyle a_{21}=8$. What is $\displaystyle a_5\ ?$Originally Posted by machilove
Given that $\displaystyle \{a_i,\ i=1,2,..\}$ is an arithmetic progression we know that $\displaystyle a_n=a_1+(n-1)d$, where $\displaystyle a_1$ is the first term, and $\displaystyle d$
is the common difference.
So:
$\displaystyle a_{13}=a_1+12d=4$
and:
$\displaystyle a_{21}=a_1+20d=8$
Subtracting the first of these from the second gives:
$\displaystyle 8d=4$, so $\displaystyle d=1/2$.
Substituting this value for $\displaystyle d$ back into one of the equations gives $\displaystyle a=-2$.
So:
$\displaystyle a_5=-2+4 \times (1/2) =0$
RonL