Question
If $a_n=3-4 n$, show that $a_1, a_2, a_3, \ldots$ form an A.P. Also find $S_{20}$.
✓
Answer
$
\begin{aligned}
& a_n=3-4 n \\
& a_1=3-4 \times 1=3-4=-1 \\
& a_2=3-4 \times 2=3-8=-5 \\
& a_3=3-4 \times 3=3-12=-9 \\
& a_4=3-4 \times 4=3-16=-13 \text { and so on }
\end{aligned}
$
Here, $a=-1, d=-5-(-1)=-5+1=-4$
Now $_1 S_{20}=\frac{n}{2}[2 a+(n-1) d]$
$
\begin{aligned}
& =\frac{20}{2}[2 \times(-1)+(20-1) \times(-4)] \\
& =10[-2+19 \times(-4)] \\
& =10[-2-76] \\
& =10 \times(-78) \\
& =-780 .
\end{aligned}
$
\begin{aligned}
& a_n=3-4 n \\
& a_1=3-4 \times 1=3-4=-1 \\
& a_2=3-4 \times 2=3-8=-5 \\
& a_3=3-4 \times 3=3-12=-9 \\
& a_4=3-4 \times 4=3-16=-13 \text { and so on }
\end{aligned}
$
Here, $a=-1, d=-5-(-1)=-5+1=-4$
Now $_1 S_{20}=\frac{n}{2}[2 a+(n-1) d]$
$
\begin{aligned}
& =\frac{20}{2}[2 \times(-1)+(20-1) \times(-4)] \\
& =10[-2+19 \times(-4)] \\
& =10[-2-76] \\
& =10 \times(-78) \\
& =-780 .
\end{aligned}
$
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