# If $a_{n}=3-4 n$, show that $a_{1}, a_{2}, a_{3}, \ldots$ form an AP. Also find $S_{20}$.

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Given:

$a_{n}=3-4 n$

To do:

We have to show that $a_{1}, a_{2}, a_{3}, \ldots$ form an AP and find $S_{20}$.

Solution:

To find $a_{1}$, we have to substitute $1$ in place of $n$ in $a_{n}=3-4n$.

This implies,

$a_{1}=a=3-4(1)$

$=3-4$

$=-1$.

To find $a_{2}$, we have to substitute $2$ in place of $n$ in $a_{n}=3-4n$.

This implies,

$a_{2}=3-4(2)$

$=3-8$

$=-5$.

To find $a_{3}$, we have to substitute $3$ in place of $n$ in $a_{n}=3-4n$.

This implies,

$a_{3}=3-4(3)$

$=3-12$

$=-9$

$a_2-a_1=-5-(-1)$

$=-5+1$

$=-4$

$a_3-a_2=-9-(-5)$

$=-9+5$

$=-4$

Here, $a_2-a_1=a_3-a_2$

Therefore,

$a_{1}, a_{2}, a_{3}, \ldots$ form an AP.

We know that,

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

$S_{20}=\frac{20}{2}[2(-1)+(20-1)(-4)]$

$=10(-2-19\times4)$

$=10(-2-76)$

$=10(-78)$

$=-780$

Updated on 10-Oct-2022 13:27:34