Question
The $17th$ term of an $AP$ exceeds its $10th$ term by $7$. Find the common difference.
✓
Answer
Let the first term and the common difference of the $AP$ be $a$ and $d$ respectively.
Given that, $a_{17}= a_{10}+7$
$ \Rightarrow a + (17 - 1) d = a + (10 - 1)d + 7 [\because a_n= a +(n - 1)d]$
$ \Rightarrow a + 16d = a + 9d + 7$
$ \Rightarrow 16d - 9d = 7$
$ \Rightarrow 7d = 7$
$ \Rightarrow d = \frac{7}{7} = 1$
Hence, the common difference is $1.$
Given that, $a_{17}= a_{10}+7$
$ \Rightarrow a + (17 - 1) d = a + (10 - 1)d + 7 [\because a_n= a +(n - 1)d]$
$ \Rightarrow a + 16d = a + 9d + 7$
$ \Rightarrow 16d - 9d = 7$
$ \Rightarrow 7d = 7$
$ \Rightarrow d = \frac{7}{7} = 1$
Hence, the common difference is $1.$
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