Question

the 17th term of an A.P exceeds it's 10th term by 7. find the common difference.​

Answer 1

Answer:

Let first term is a and common difference is d

and given that

T17=T10+7

a+16d=a+9d+7

a+16d-a-9d=7

7d=7

on dividing by 7

hence d=1

Answer:

common difference is 1.

Step-by-step explanation:

Answer 2

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

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

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

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