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A man accepts a position with an initial salary of $₹ 5200$ per month. It is understood that he will receive an automatic increase of $₹ 320$ in the every next month and each month thereafter.
(i) Find his salary for the tenth month.
(ii) What is his total earnings during the first year?
(i) Find his salary for the tenth month.
(ii) What is his total earnings during the first year?
Solution:
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Verified Answer
Since, the man get a fixed increment of $₹ 320$ each month. Therefore, this forms an AP whose First term $=5200$ and Common difference $(d)=320$
(i) Salary for tenth month is calculated as:
Here, $n=10$,
$$
\begin{aligned}
&\therefore a_{10}=a+(n-1) d \\
&\Rightarrow a_{10}=5200+(10-1) \times 320 \\
&\Rightarrow a_{10}=5200+9 \times 320 \\
&\Rightarrow a_{10}=5200+2880 \\
&\Rightarrow a_{10}=8080
\end{aligned}
$$
(ii) Total earning during the first year is calculated as: Here, $\quad n=12$,
$$
\begin{aligned}
&\therefore S_{12}=\frac{12}{2}[2 \times 5200+(12-1) 320] \\
&=6[10400+11 \times 320] \\
&=6[10400+3520]=6 \times 13920=83520
\end{aligned}
$$
(i) Salary for tenth month is calculated as:
Here, $n=10$,
$$
\begin{aligned}
&\therefore a_{10}=a+(n-1) d \\
&\Rightarrow a_{10}=5200+(10-1) \times 320 \\
&\Rightarrow a_{10}=5200+9 \times 320 \\
&\Rightarrow a_{10}=5200+2880 \\
&\Rightarrow a_{10}=8080
\end{aligned}
$$
(ii) Total earning during the first year is calculated as: Here, $\quad n=12$,
$$
\begin{aligned}
&\therefore S_{12}=\frac{12}{2}[2 \times 5200+(12-1) 320] \\
&=6[10400+11 \times 320] \\
&=6[10400+3520]=6 \times 13920=83520
\end{aligned}
$$
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