The 17^{th} term of an AP exceeds its 10^{th} | vedclass

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Question

(D) We know that for an $AP$,the $n^{th}$ term is given by the formula: $a_n = a + (n - 1)d$.For the $17^{th}$ term:$a_{17} = a + (17 - 1)d = a + 16d$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) We know that for an $AP$,the $n^{th}$ term is given by the formula: $a_n = a + (n - 1)d$.For the $17^{th}$ term:$a_{17} = a + (17 - 1)d = a + 16d$.For the $10^{th}$ term:$a_{10} = a + (10 - 1)d = a + 9d$.According to the problem,the $17^{th}$ term exceeds the $10^{th}$ term by $7$:$a_{17} - a_{10} = 7$.Substituting the expressions:$(a + 16d) - (a + 9d) = 7$.Simplifying the equation:$a + 16d - a - 9d = 7$.$7d = 7$.Dividing both sides by $7$:$d = 1$.Therefore,the common difference is $1$.

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

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