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12th term of an \( \mathrm{AP} \) is 4 and its 20th term is \( -20 \). Find the \( n \) th term of that \( A P \).
Given:
12th term of an \( \mathrm{AP} \) is 4 and its 20th term is \( -20 \).
To do:
We have to find the \( n \) th term of that \( A P \).
Solution:
Let the first term of the A.P. be $a$ and the common difference be $d$.
We know that,
nth term of an A.P. $a_n=a+(n-1)d$
Therefore,
$a_{12}=a+(12-1)d$
$4=a+11d$......(i)
$a_{20}=a+(20-1)d$
$-20=a+19d$.......(ii)
Subtracting (i) from (ii), we get,
$-20-4=a+19d-(a+11d)$
$-24=a-a+19d-11d$
$-24=8d$
$d=\frac{-24}{8}$
$d=-3$
$\Rightarrow 4=a+11d$ (From (i))
$4=a+11(-3)$
$a=4+33$
$a=37$
$\Rightarrow a_n=a+(n-1)d$
$=37+(n-1)(-3)$
$=37-3n+3$
$=40-3n$
The $n^{th}$ term of the AP is $40-3n$.
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