Question

In an A.Pi)  Given $$a=2,\,d=8,\,\,Sn=90$$, Find n & $${a_9}$$ii) Given $$d = 5,{S_9} = 75$$, find $$a\,\& \,{a_9}$$iii) Given $$a = 7,\,\,{a_{13}} = 35$$, find $$d\,\& \,{S_{13}}$$

Solution

(i)Here $$a=2,d=8,S_n=90$$We know $$S_n=\dfrac{n}{2}[2a+(n-1)d]$$$$90=\dfrac{n}{2}[2\cdot 2+(n-1)8]$$$$\Rightarrow 180=n[4+8n-8]$$$$\Rightarrow 180=8n^2-4n$$$$\Rightarrow 8n^2-4n-180=0$$$$\Rightarrow 2n^2-n-45=0$$$$\Rightarrow (n-5)(2n+9)=0$$$$\Rightarrow n=5,-\dfrac{9}{2}$$$$\Rightarrow n=5$$ (since n cannot be negative)Therefore $$a_n=a+4d=2+4(8)=2+32=34$$(ii)Here $$d=5, S_9=75$$We know $$S_n=\dfrac{n}{2}[2a+(n-1)d]$$$$S_9=\dfrac{9}{2}[2a+(8)d]$$$$\Rightarrow 75=\dfrac{9}{2}[2a+8(5)]$$$$\Rightarrow 75=\dfrac{9}{2}[2a+40]$$$$\Rightarrow 75=9[a+20]$$$$\Rightarrow 75=9a+180$$$$\Rightarrow 9a=-105$$$$\Rightarrow a=-\dfrac{35}{3}$$Therefore $$a_9=a+8d=-\dfrac{35}{3}+8(5)=\dfrac{85}{3}$$(iii)Here $$a=7,a_{13}=35$$We know $$a_n=a+(n-1)d$$$$\Rightarrow a_{13}=7+(13-1)d$$$$\Rightarrow 35=7+12d$$$$\Rightarrow 28=12d$$$$\Rightarrow d=\dfrac{7}{3}$$We know $$S_n=\dfrac{n}{2}[2a+(n-1)d]$$$$S_{13}=\dfrac{13}{2}\left[2(7)+(13-1)\dfrac{7}{3}\right]$$       $$=\dfrac{13}{2}\left[14+12\times \dfrac{7}{3}\right]$$       $$=\dfrac{13}{2}[14+28]$$       $$=\dfrac{13}{2}(42)$$       $$=13\times 21$$$$S_{13}=273$$Mathematics

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