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The sum of first $n$ terms of an AP is $5n^2 + 3n$. If its $n^{th}$ term is $168$, find $n$. Also, find the $20^{th}$ term of the AP.
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Given $S_n = 5n^2 + 3n$.
For $n=1$, $S_1 = 5(1)^2 + 3(1) = 5+3 = 8$. So, $a_1 = 8$.
For $n=2$, $S_2 = 5(2)^2 + 3(2) = 5(4) + 6 = 20+6 = 26$.
$S_2 = a_1 + a_2 \Rightarrow 26 = 8 + a_2 \Rightarrow a_2 = 18$.
Common difference $d = a_2 - a_1 = 18 - 8 = 10$.
Now, $a_n = 168$.
$a_n = a_1 + (n-1)d$
$168 = 8 + (n-1)10$
$160 = (n-1)10$
$16 = n-1 \Rightarrow n = 17$.
To find $20^{th}$ term, $a_{20} = a_1 + (20-1)d = 8 + 19(10) = 8 + 190 = 198$.
For $n=1$, $S_1 = 5(1)^2 + 3(1) = 5+3 = 8$. So, $a_1 = 8$.
For $n=2$, $S_2 = 5(2)^2 + 3(2) = 5(4) + 6 = 20+6 = 26$.
$S_2 = a_1 + a_2 \Rightarrow 26 = 8 + a_2 \Rightarrow a_2 = 18$.
Common difference $d = a_2 - a_1 = 18 - 8 = 10$.
Now, $a_n = 168$.
$a_n = a_1 + (n-1)d$
$168 = 8 + (n-1)10$
$160 = (n-1)10$
$16 = n-1 \Rightarrow n = 17$.
To find $20^{th}$ term, $a_{20} = a_1 + (20-1)d = 8 + 19(10) = 8 + 190 = 198$.