Question

in an A.P 17th term is 7 more than its 10th term find the commom difference ​

Answer 1

a17 = 7+a10

a+16 d =7+a+9d

16d-9d=7

7d = 7

d=7/7

d=1

Answer 2

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

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

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

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