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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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.

Similar Questions

In the following arithmetic progression $(AP)$,find the missing terms in the boxes: $\square, 13, \square, 3$.

Choose the correct choice in the following and justify: $30^{th}$ term of the $AP: 10, 7, 4, \ldots$ is

For a given $A.P.$,the $11^{th}$ term is $38$ and the $16^{th}$ term is $73$. Find the $31^{st}$ term of the $A.P.$

For the finite $A.P.$ $3, 5, 7, \ldots, 201$,find the $12^{th}$ term from the end.

Which term of the $A.P.$ $5 \frac{1}{2}, 11, 16 \frac{1}{2}, \ldots$ is $550$ (in $^{th}$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module