Question

the 17 term of the an ap exceed its term by7. find the common difference.

Answer 1

Let a be the first term and d be the common difference of A.P.

17th term of A.P t17 = a + 16d.

10th term of A.P is t10 = a + 9d.

Given that 17th term of an A.P exceeds its 10th term by 7.

a + 16d = a + 9d + 7

7d = 7

∴ d = 1

Answer 2

$\bold{\huge\purple{\boxed{{{QUESTION}}}}}$

The 17th term of an AP exceed is 10th term by 7. find the common differnce.

$\bold{\huge\red{\boxed{{{ANSWER}}}}}$

$Let \: a \: be \: the \: first \: term \: and \\ \: d \: be \: the \: common \: diffrence \: of \: the \: given \: AP \\ \\ Now, \: according \: to \: the \: question \: a17 = a10 + 7 \\ = > a17 - a10 = 7 \\ = > a + (17 - 1)d - a + (10 - 1)d = 7 \\ ( an = a + (n - 1)d) \\ \\ = > \: \: (a + 16d) - (a + 9d) = 7 \\ = > \: 7d \: = 7 \\ = > \: \: d = 1 \\ \\ Hence, \: the \: common \: diffrence \: of \: this \: ap \: is \: 1.$