Question
In an A.P. $17^{\text {th }}$ term is $7$ more than its $10^{\text {th }}$ term. Find the common difference.
✓
Answer
Given: $t_{17}=7+t_{10}$
$\text { In } t _{17}, n =17$
$\text { In } t _{10}, n =10$
By using $n ^{\text {th }}$ term of an A.P. formula,
$t_n=a+(n-1) d$
where $n =$ no. of terms
$a =\text { first term }$
$d=$ common difference
$t _{ n }= n ^{\text {th }} \text { term }$
Thus, on using formula in eq. (1) we get,
$\Rightarrow a+(17-1) d=7+(a+(10-1) d)$
$\Rightarrow a+16 d=7+(a+9 d)$
$\Rightarrow a+16 d-a-9 d=7$
$\Rightarrow 7 d=7$
$\Rightarrow d=\frac{7}{7}=1$
Thus, common difference “d” = 1
$\text { In } t _{17}, n =17$
$\text { In } t _{10}, n =10$
By using $n ^{\text {th }}$ term of an A.P. formula,
$t_n=a+(n-1) d$
where $n =$ no. of terms
$a =\text { first term }$
$d=$ common difference
$t _{ n }= n ^{\text {th }} \text { term }$
Thus, on using formula in eq. (1) we get,
$\Rightarrow a+(17-1) d=7+(a+(10-1) d)$
$\Rightarrow a+16 d=7+(a+9 d)$
$\Rightarrow a+16 d-a-9 d=7$
$\Rightarrow 7 d=7$
$\Rightarrow d=\frac{7}{7}=1$
Thus, common difference “d” = 1
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