Question

In an A.P
$\begin{array}{c}\text{(i) given}{a}_3=15,S_{10}=125,\text{find}d\text{and}{a}_{10}\\ \text{(ii) given}d=5,S_9=75,\text{find}a\text{and}{a}_9\text{.}\end{array}$

Answer 1

i)$a_3=15,S_{10}=125a_n=a_1+(n-1)d{a}_3=a_1+(n-1)d15=a+(3-1)d\Rightarrow a+2d=15\Rightarrow d=\frac{(15-a)}{2}{S}_n=\frac{n}{2}[2a+(n-1\left)d\right]S_{10}=\frac{10}{2}[2a+(10-1\left)d\right]=125\Rightarrow \frac{125}{25}=5[2a+9d]\Rightarrow 25=2a+9\left(\frac{15-a}{2}\right)\Rightarrow 50=4a+135-9a\Rightarrow 5a=135-50=85\Rightarrow a=\frac{85}{5}=1715=a+2d\Rightarrow 15=17+2d15-17=2d\Rightarrow -2=2d\Rightarrow d=-1\therefore a=17,d=-1a_{10}=a+(10-1)d=17+9(-1)=17-9=8$

ii)$d=5,S_9=75S_9=\frac{9}{2}[2a+(n-1\left)d\right]\Rightarrow 75=\frac{9}{2}[2a+(9-1\left)5\right]\Rightarrow 150=9(2a+40)\Rightarrow 150=18a+360\Rightarrow 150-360=18a\Rightarrow -210=18a\Rightarrow a=\frac{-210}{18}=\frac{-35}{3}{a}_9=a+(9-1)da+8d=\frac{-35}{3}+8\times 5=\frac{-35}{3}+40=\frac{120-35}{3}=\frac{85}{3}\therefore a_9=\frac{85}{3}$