
Write the AP whose ${{\text{n}}^{\text{th}}}$ term is given by ${{a}_{n}}\text{ = 9 }-\text{ 5n}$.
Answer
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Hint: Use the general mth term of an AP which is given by ${{a}_{m}}\text{ = a + }\left( \text{m }-\text{ 1} \right)\text{d}$, where ‘a’ is the first term of the arithmetic progression and ‘d’ is its common difference. Now compare the nth term given in the question to any general nth term of an AP, and thus by comparison method find the values of the first term and the common difference of the AP.
Complete step-by-step answer:
We know, the general mth term of an AP is given by, ${{a}_{m}}\text{ = a + }\left( \text{m }-\text{ 1} \right)\text{d}$ where ‘a’ is the first term of the arithmetic progression and ‘d’ is its common difference.
Now, it is given that for an AP, the nth term is given by the expression ${{a}_{n}}\text{ = 9 }-\text{ 5n}$. Thus, we need to compare this expression with the nth term of any general AP.
Thereby, putting m = n in the expression of the general term of the AP, we get,
$\begin{align}
& {{a}_{n}}\text{ = 9 }-\text{ 5n} \\
& \Rightarrow \text{ a + }\left( \text{n }-\text{ 1} \right)\text{d = 9 }-\text{ 5n} \\
& \therefore \text{ }\left( \text{a }-\text{ d} \right)\text{ + d}\cdot \text{n = 9 }-\text{ 5n} \\
\end{align}$
Thus, from the above equation, we compare the coefficients of 1 and n, and equate them by the method of identity. Hence, we get,
\[\begin{align}
& \text{a }-\text{ d = 9 }....\text{(i)} \\
& \text{d = }-\text{5 }....\text{(ii)} \\
\end{align}\]
From equation (ii), we get the value of the common difference ‘d’ as –5.
Putting the value of ‘d’ in equation (i), we get,
$\begin{align}
& \text{a }-\text{ }\left( -5 \right)\text{ = 9} \\
& \therefore \text{ a = 9 }-\text{ 5 = 4} \\
\end{align}$
Thus, the first term of the AP is 4 and the common difference is –5. So, we write the general mth term of the given AP as ${{a}_{m}}\text{ = 4 }-\text{ 5}\left( \text{m }-\text{ 1} \right)$.
Note: Arithmetic Progression is a sequence where the terms of the sequence are either increasing or decreasing, with the difference between the adjacent terms being constant. Here, in this problem, as the common difference is negative, it is an example of decreasing AP.
Complete step-by-step answer:
We know, the general mth term of an AP is given by, ${{a}_{m}}\text{ = a + }\left( \text{m }-\text{ 1} \right)\text{d}$ where ‘a’ is the first term of the arithmetic progression and ‘d’ is its common difference.
Now, it is given that for an AP, the nth term is given by the expression ${{a}_{n}}\text{ = 9 }-\text{ 5n}$. Thus, we need to compare this expression with the nth term of any general AP.
Thereby, putting m = n in the expression of the general term of the AP, we get,
$\begin{align}
& {{a}_{n}}\text{ = 9 }-\text{ 5n} \\
& \Rightarrow \text{ a + }\left( \text{n }-\text{ 1} \right)\text{d = 9 }-\text{ 5n} \\
& \therefore \text{ }\left( \text{a }-\text{ d} \right)\text{ + d}\cdot \text{n = 9 }-\text{ 5n} \\
\end{align}$
Thus, from the above equation, we compare the coefficients of 1 and n, and equate them by the method of identity. Hence, we get,
\[\begin{align}
& \text{a }-\text{ d = 9 }....\text{(i)} \\
& \text{d = }-\text{5 }....\text{(ii)} \\
\end{align}\]
From equation (ii), we get the value of the common difference ‘d’ as –5.
Putting the value of ‘d’ in equation (i), we get,
$\begin{align}
& \text{a }-\text{ }\left( -5 \right)\text{ = 9} \\
& \therefore \text{ a = 9 }-\text{ 5 = 4} \\
\end{align}$
Thus, the first term of the AP is 4 and the common difference is –5. So, we write the general mth term of the given AP as ${{a}_{m}}\text{ = 4 }-\text{ 5}\left( \text{m }-\text{ 1} \right)$.
Note: Arithmetic Progression is a sequence where the terms of the sequence are either increasing or decreasing, with the difference between the adjacent terms being constant. Here, in this problem, as the common difference is negative, it is an example of decreasing AP.
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