
In an AP, the 17th term is 7 more than its 10th term. Find the common difference.
Answer
590.1k+ views
Hint: In order to solve this question, we should know about the concept of arithmetic progression. And we should know that the nth term of an AP can be calculated by using the formula, ${{a}_{n}}=a+\left( n-1 \right)d$, where a is the first term, d is the common difference and n is the number of terms. By using these concepts, we will solve this question.
Complete step-by-step answer:
In this question, we have been asked to find the common difference of an AP whose 17th term is 7 more than its 10th term. To solve this question, we should know that AP or arithmetic progression is a progression of numbers, such that the difference between the consecutive terms is constant and that is called the common difference. For example, consider the AP, a, a+d, a+2d, a+3d,…… a+(n-1)d. Here, a represents the first term, d represents the common difference and n is the number of terms.
Now, we know that the n th term of an AP is calculated by using the formula, ${{a}_{n}}=a+\left( n-1 \right)d$. So, we can say that the 10th term can be written as, ${{a}_{10}}=a+\left( 10-1 \right)d=a+9d$ and we can write the 17th term as, ${{a}_{17}}=a+\left( 17-1 \right)d=a+16d$.
Now, according to the condition given question, we can write it as follows.
${{a}_{17}}=7+{{a}_{10}}$
Now, we will put the values of ${{a}_{17}}$ and ${{a}_{10}}$ in the above equation. So, we will get,
$a+16d=7+a+9d$
Now, we will keep the terms containing the variable d on one side and the rest all on the other side. So, we get,
$16d-9d=7+a-a$
Now, we know that like terms show algebraic addition. So, we can write,
7d = 7
d = 1
Hence, we can say that the common difference of the AP whose 17th term is 7 more than its 10th term is 1.
Note: We could have also solved this question in an alternative way, as we know that the number of terms from 10th term to the 17th term is 7 and the difference between these values is also 7. So, we can say that the common difference of $AP=\dfrac{7}{7}=1$. But this can be a confusing way of solving for larger numbers and there are possibilities that we might make a mistake using this method. So, it is always better to use the formula that is ${{a}_{n}}=a+\left( n-1 \right)d$ to solve this question.
Complete step-by-step answer:
In this question, we have been asked to find the common difference of an AP whose 17th term is 7 more than its 10th term. To solve this question, we should know that AP or arithmetic progression is a progression of numbers, such that the difference between the consecutive terms is constant and that is called the common difference. For example, consider the AP, a, a+d, a+2d, a+3d,…… a+(n-1)d. Here, a represents the first term, d represents the common difference and n is the number of terms.
Now, we know that the n th term of an AP is calculated by using the formula, ${{a}_{n}}=a+\left( n-1 \right)d$. So, we can say that the 10th term can be written as, ${{a}_{10}}=a+\left( 10-1 \right)d=a+9d$ and we can write the 17th term as, ${{a}_{17}}=a+\left( 17-1 \right)d=a+16d$.
Now, according to the condition given question, we can write it as follows.
${{a}_{17}}=7+{{a}_{10}}$
Now, we will put the values of ${{a}_{17}}$ and ${{a}_{10}}$ in the above equation. So, we will get,
$a+16d=7+a+9d$
Now, we will keep the terms containing the variable d on one side and the rest all on the other side. So, we get,
$16d-9d=7+a-a$
Now, we know that like terms show algebraic addition. So, we can write,
7d = 7
d = 1
Hence, we can say that the common difference of the AP whose 17th term is 7 more than its 10th term is 1.
Note: We could have also solved this question in an alternative way, as we know that the number of terms from 10th term to the 17th term is 7 and the difference between these values is also 7. So, we can say that the common difference of $AP=\dfrac{7}{7}=1$. But this can be a confusing way of solving for larger numbers and there are possibilities that we might make a mistake using this method. So, it is always better to use the formula that is ${{a}_{n}}=a+\left( n-1 \right)d$ to solve this question.
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