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