Question
In an $A.P.$ (with usual notations) : given $a = 7, a_{13} = 35$, find d and $S_{13}$
✓
Answer
$
\begin{aligned}
& a=7, a_{13}=35 \\
& a_n=a+(n-1) d \\
& 35=7+(13-1) d \\
& \Rightarrow 35-7=12 d \\
& \Rightarrow 28=12 d \\
& \Rightarrow d=\frac{28}{12} \\
& =\frac{7}{3} \\
& =2 \frac{1}{3}
\end{aligned}
$
and
$
\begin{aligned}
& S_{13}=\frac{n}{2}[2 a+(n-1) d] \\
& =\frac{13}{2}\left[2 \times 7+(13-1) \times \frac{7}{3}\right] \\
& =\frac{13}{2}[14+28] \\
& =\frac{13}{2} \times(42) \\
& =13 \times 21 \\
& =273 .
\end{aligned}
$
\begin{aligned}
& a=7, a_{13}=35 \\
& a_n=a+(n-1) d \\
& 35=7+(13-1) d \\
& \Rightarrow 35-7=12 d \\
& \Rightarrow 28=12 d \\
& \Rightarrow d=\frac{28}{12} \\
& =\frac{7}{3} \\
& =2 \frac{1}{3}
\end{aligned}
$
and
$
\begin{aligned}
& S_{13}=\frac{n}{2}[2 a+(n-1) d] \\
& =\frac{13}{2}\left[2 \times 7+(13-1) \times \frac{7}{3}\right] \\
& =\frac{13}{2}[14+28] \\
& =\frac{13}{2} \times(42) \\
& =13 \times 21 \\
& =273 .
\end{aligned}
$
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