Question

The ${17^{th}}$ term of an A.P. exceeds its ${10^{th}}$ term by 7. Find the common difference.

Answer 1

We will write the ${17^{th}}$ term and ${10^{th}}$ term of the A.P. using the expression of ${n^{th}}$ term of an A.P which is given by ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term and $d$ is the common difference of the A.P. Next, form an equation according to the given condition and solve for the value of $d$.

Complete step-by-step answer:
The ${n^{th}}$ term of an A.P is given by ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term and $d$ is the common difference of the A.P.
The ${17^{th}}$ term of an A.P will be given by
${a_{17}} = a + \left( {17 - 1} \right)d \\\ \Rightarrow {a_{17}} = a + 16d \\\$
And the ${10^{th}}$ term is
${a_{10}} = a + \left( {10 - 1} \right)d \\\ \Rightarrow {a_{10}} = a + 9d \\\$
We are given that ${17^{th}}$ term of an A.P. exceeds its ${10^{th}}$ term by 7.
This implies, ${a_{17}} = {a_{10}} + 7$
On substituting the value of ${a_{10}}$ and ${a_{17}}$ in the above equation, we get,
$a + 16d = a + 9d + 7 \\\ \Rightarrow 16d = 9d + 7 \\\ \Rightarrow 7d = 7 \\\$
Divide both sides by 7
$d = 1$
Hence, the common difference of A.P is 1.

Note: An A.P. is of the form, $a,a + d,a + 2d,a + 3d,........a + \left( {n - 1} \right)d$, where $a$ is the first term, $d$ is the common difference and $n$ is the number of terms of the A.P. Here, we cannot solve for the value of $a$. For solving two variables, we need two equations involving those variables.