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Question

In an A.P
i)  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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