Question

Of $a,l,d,n,Sn$ determine the ones which are missing for the following arithmetic progressions :
$\begin{array}{cc}\left(i\right)a=-2,d=5,S_n=568& \left(ii\right)l=8,n=8,S_8=-20\\ \left(ii\right)d=\frac{2}{3},l=10,n=20& \left(iv\right)a=2,l=29,S_n=155\\ \left(v\right)a=8,l=62,S_n=210& \left(vi\right)l=4,d=2,S_n=-14\end{array}$

Answer 1

$i)a=-2,d=5,S_n=568$

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)=568$

$\Rightarrow \frac{n}{2}(-4+(n-1\left)5\right)=568$

$\Rightarrow n(5n-9)=1136$

$\Rightarrow 5n^2-9n-1136=0$

$\Rightarrow n=\frac{9\pm \sqrt{81+4\times 5\times 1136}}{2\times 5}$

$=\frac{9\pm 151}{10}=16$ $-14.2$

$n=16$

$l=a+(n-1)d=-2+15\times 5+23$

$ii)l=8,n=8,S_8=-20$

$l=a+(n-1)d\Rightarrow a+7d=8$

$S_8=\frac{8}{2}(a+l)=4(a+8)=-20$

$\Rightarrow a=-13\Rightarrow d=3$

$iii)d=\frac{2}{3},l=10,n=20$

$l=10=a+(n-1)d=a+19\times \frac{2}{3}$

$\Rightarrow 10=a+19\times \frac{2}{3}$

$\Rightarrow a=10-\frac{38}{3}=-\frac{8}{3}$

$S_n=\frac{n}{2}(a+l)=\frac{20}{2}(-\frac{8}{3}+10)$

$=10\left(\frac{22}{3}\right)=\frac{220}{3}$

$iv)a=2,l=29,S_n=155$

$Sn=\frac{n}{2}(a+l)\Rightarrow \frac{n}{2}(2+29)=155$

$\Rightarrow n\left(31\right)=310\Rightarrow n=10$

$l=a+(n-1)d\Rightarrow 29=2+9d$

$\Rightarrow d=3$

$v)a=8,l=62,S_n=210$

$S_n=\frac{n}{2}(a+l)=\frac{n}{2}(8+62)$

$210=\frac{n}{2}\left(70\right)\Rightarrow n=6$

$l=a+(n-1)d\Rightarrow 62=8+5d$

$\Rightarrow d=10.8$

$vi)l=4,d=2,S_n=-14$

$l=4=a+(n-1)d=a+2(n-1)$ ...(1)

$S_n=-14=\frac{n}{2}(a+l)=\frac{n}{2}(a+4)$

$\Rightarrow n(a+4)=-28$ ...(2)

$\Rightarrow 4=a+2n-2\Rightarrow a+2n=6$

$\Rightarrow a=6-2n$

put $a=6-2n$ in ....(2)

$n(6-2n+4)=-28$

$\Rightarrow n(10-2n)=-28$

$\Rightarrow 2n^2-10n-28=0$

$\Rightarrow n^2-5n-14=0$

$\Rightarrow (n+2)(n-7)=0$

$\Rightarrow n=-2,7\Rightarrow n=7$

$a=6-2n=6-14=-8$