If $ a_{n}=3-4 n $, show that $ a_{1}, a_{2}, a_{3}, \ldots $ form an AP. Also find $ S_{20} $.
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,
$S_{n}=\frac{n}{2}[2a+(n-1)d]$
$S_{20}=\frac{20}{2}[2(-1)+(20-1)(-4)]$
$=10(-2-19\times4)$
$=10(-2-76)$
$=10(-78)$
$=-780$
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