Question

Show that $a_1,a_2,...,a_n$ form an AP where $a_n$ is defined as below :
(i) $a_n=3+4n$
(ii) $a_n=9-5n$
Also find the sum of the first $15$ terms in each case.

Answer 1

(i) $a_n=3+4n$ given

$a_1=3+4\left(1\right)=7$

$a_2=3+4\left(2\right)=11$

$a_2=3+4\left(3\right)=15$

$\therefore d=a_3-a_1=11-7=4$

Here $a=7$, $d=4$ and $n=15$

By using$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ we have,

$S_{15}=\frac{15}{2}[2\times 7+(15-1\left)4\right]$

$=\frac{15}{2}(14+56)$

$=\frac{15}{2}\times 70=525$

(ii) $a_n=9-5n$ given

$a_1=9-5\left(1\right)=9-5=4$

$a_2=9-5\left(2\right)=9-10=-1$

$a_3=9-5\left(3\right)=9-15=-6$

$\therefore d=a_2-a_1=-6-(-1)=-5$

Here $a=4,d=-5$ and $n=15$

By using $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ we have,

$S_{15}=\frac{15}{2}[2\times 4+(15-1\left)\right(-5\left)\right]$

$=\frac{15}{2}(8-70)$

$=\frac{15}{2}\times (-62)=-465$