Question

(i) the sum fo the first n terms of an AP is $(\frac{5n^2}{2}+\frac{3n}{2})$. Find hte nth term and the 20th term of this AP.
(ii) The sum of the first n terms of an AP is $(\frac{3n^2}{2}+\frac{5n}{2})$. Find its nth term and the 25th term.

Answer 1

(i) Sum of n terms, $S_n=\frac{5n^2}{2}+\frac{3n}{2}$

$S_1=\frac{5(1{)}^2}{2}+\frac{3\left(1\right)}{2}=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4=a$

$S_1=\frac{5(2{)}^2}{2}+\frac{3\left(2\right)}{2}=\frac{20}{2}+\frac{6}{2}=10+3=13$

$a_2=S_2-S_1=13-4=9$

$d=a_2-a_1=9-4=5$

$a_n=a+(n-1)d=4+(n-1)5=4+5n-5=5n-1$

$a_{20}=5n-1=5\times 20-1=100-1=99$

(ii) Sum of n terms, $S_n=\frac{3n^2}{2}+\frac{5n}{2}$

$S_1=\frac{3(1{)}^2}{2}+\frac{5\left(1\right)}{2}=\frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4=a$

$S_1=\frac{3(2{)}^2}{2}+\frac{3\left(2\right)}{2}=\frac{12}{2}+\frac{10}{2}=6+5=11$

$a_2=S_2-S_1=11-4=7$

$d=a_2-a_1=7-4=3$

$a_n=a+(n-1)d=4+(n-1)3=4+3n-3=3n+1$

$a_{25}=3n+1=3\times 25+1=75+1=76$