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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Updated on: 10-Oct-2022

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