Question

sn=4n²+5n.a) what is the common different.b)X1?and S25​

Answer 1

Given:-

$\\ \rm\longmapsto S_n=4n^2+5n$

To find:-

$\\ \rm\longmapsto S_{25}$

$\\ \rm\longmapsto x_n$

$\\ \rm\longmapsto d$

Solution:-

$\\ \rm\longmapsto S_n=4n^2+5n$

Now

$\\ \rm\longmapsto S_{(n-1)}=4(n-1)^2+5(n-1)$

$\\ \rm\longmapsto S_{(n-1)}=4(n^2-2n+1)+5n-5$

$\\ \rm\longmapsto S_{(n-1)}=4n^2-8n+4+5n-5$

$\\ \rm\longmapsto S_{(n-1)}=4n^2-8n+5n+4-5$

$\\ \rm\longmapsto S_{(n-1)}=4n^2-3n-1$

We know that

$\boxed{\sf x_n=S_n-S_{n-1}}$

$\\ \rm\longmapsto x_n=4n^2+5n-(4n^2-3n-1)$

$\\ \rm\longmapsto x_n=4n^2+5n-4n^2+3n+1$

$\\ \rm\longmapsto x_n=4n^2-4n^2+5n+3n+1$

$\\ \rm\longmapsto x_n=8n+1$

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We have now

$\boxed{\sf x_n=8n+1}$

$\\ \rm\longmapsto x_1=8(1)+1$

$\\ \rm\longmapsto x_1=8+1$

$\\ \rm\longmapsto \underline{\boxed{\bf{x_1=9}}}$

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$\\ \rm\longmapsto x_2=8(2)+1$

$\\ \rm\longmapsto x_2=16+1$

$\\ \rm\longmapsto x_2=17$

We know

$\boxed{\sf d=x_2-x_1}$

$\\ \rm\longmapsto d=17-9$

$\\ \rm\longmapsto\underline{\boxed{\bf{d=8}}}$

____________________________

• a=9
• d=8

We know

$\boxed{\sf S_n=\dfrac{n}{2}\left\{2a+(n-1)d\right\}}$

$\\ \rm\longmapsto S_{25}=\dfrac{25}{2}\left\{2(9)+(25-1)8\right\}$

$\\ \rm\longmapsto S_{25}=\dfrac{25}{2}\left\{18+24\times 8\right\}$

$\\ \rm\longmapsto S_{25}=\dfrac{25}{2}\left\{18+192\right\}$

$\\ \rm\longmapsto S_{25}=\dfrac{25}{\cancel{2}}\times \cancel{210}$

$\\ \rm\longmapsto S_{25}=25\times 105$

$\\ \rm\longmapsto \underline{\boxed{\bf{S_{25}=2625}}}$