(a)OR(b)

Which term of the A.P. $\displaystyle -\frac{11}{2},-3,-\frac{1}{2}, \ldots$ is $\displaystyle \frac{49}{2}$ ?

Here \(\displaystyle a=\frac{-11}{2}, d=\frac{5}{2}, a_{n}=\frac{49}{2}\) \(\displaystyle \frac{49}{2}=\frac{-11}{2}+(n-1) \frac{5}{2}\) \[\Rightarrow n=13 \] Or Find a and b so that the numbers a, $\displaystyle 7$, b, $\displaystyle 23$ are in A.P. Numbers are in AP Therefore, \(\displaystyle a+b=14\) and \(\displaystyle 2 b=30\) \[\Rightarrow b=15, \quad a=-1 \]