
${{30}^{th}}$ term of AP: 10, 7, 4 . . . . . . . is
\[\begin{align}
& A.97 \\
& B.77 \\
& C.-77 \\
& D.-87 \\
\end{align}\]
Answer
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Hint: In this question, we are given an arithmetic progression and we have to find ${{30}^{th}}$ term of the given AP. For this, we will first analyze the given arithmetic progression to find the value of the first term of arithmetic progression and name it as 'a'. After that, we will find the common difference of the arithmetic progression by subtracting the first term from the second term. Since, it is already given as arithmetic progression, so the difference between third term and second term will be the same as the difference between second term and first term. Common difference will be given name as 'd'. Using value of ‘a’ and ‘d’ in the formula ${{a}_{n}}=a+\left( n-1 \right)d$ we will find ${{30}^{th}}$ term of the arithmetic progression by putting n = 30. Here, ${{a}_{n}}$ is the required answer.
Complete step by step answer:
Here we are given arithmetic progression as 10, 7, 4 . . . . . . . . . . . .
We need to find ${{30}^{th}}$ term of the given arithmetic progression so we will first find the first term and common difference of the arithmetic progression.
As we can see first term of AP 10, 7, 4. . . . . . . . . . . . . is 10. So let us name it as a. Hence, a = 10.
Now, the common difference will be given by difference between second term and first term. So, the common difference is equal to 7-10 = -3. Let us name it as d. So, d = -3.
We have to calculate ${{30}^{th}}$ term, so n will be 30. Since the formula for finding ${{n}^{th}}$ term of AP is given by ${{a}_{n}}=a+\left( n-1 \right)d$. So putting values of a, d and n for given arithmetic progression we get:
\[\begin{align}
& {{a}_{30}}=10+\left( 30-1 \right)\left( -3 \right) \\
& \Rightarrow 10+\left( 29 \right)\left( -3 \right) \\
& \Rightarrow 10-\left( 29 \right)\left( 3 \right) \\
& \Rightarrow 10-87 \\
& \Rightarrow -77 \\
\end{align}\]
Hence, ${{30}^{th}}$ term of arithmetic progression 10, 7, 4 . . . . . . . . . is -77.
So, the correct answer is “Option C”.
Note: Students should note that common differences can be negative too, so they do not make the mistake of subtracting second term from first term. Students can also subtract second term from third term to find the required common difference. Students should not get confused between n and ${{a}_{n}}$. ${{a}_{n}}$ represents the ${{n}^{th}}$ term in an AP.
Complete step by step answer:
Here we are given arithmetic progression as 10, 7, 4 . . . . . . . . . . . .
We need to find ${{30}^{th}}$ term of the given arithmetic progression so we will first find the first term and common difference of the arithmetic progression.
As we can see first term of AP 10, 7, 4. . . . . . . . . . . . . is 10. So let us name it as a. Hence, a = 10.
Now, the common difference will be given by difference between second term and first term. So, the common difference is equal to 7-10 = -3. Let us name it as d. So, d = -3.
We have to calculate ${{30}^{th}}$ term, so n will be 30. Since the formula for finding ${{n}^{th}}$ term of AP is given by ${{a}_{n}}=a+\left( n-1 \right)d$. So putting values of a, d and n for given arithmetic progression we get:
\[\begin{align}
& {{a}_{30}}=10+\left( 30-1 \right)\left( -3 \right) \\
& \Rightarrow 10+\left( 29 \right)\left( -3 \right) \\
& \Rightarrow 10-\left( 29 \right)\left( 3 \right) \\
& \Rightarrow 10-87 \\
& \Rightarrow -77 \\
\end{align}\]
Hence, ${{30}^{th}}$ term of arithmetic progression 10, 7, 4 . . . . . . . . . is -77.
So, the correct answer is “Option C”.
Note: Students should note that common differences can be negative too, so they do not make the mistake of subtracting second term from first term. Students can also subtract second term from third term to find the required common difference. Students should not get confused between n and ${{a}_{n}}$. ${{a}_{n}}$ represents the ${{n}^{th}}$ term in an AP.
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