Question

The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference. ​

Answer 1

Answer:

1

Step-by-step explanation:

According to the question,

a17 = a10 + 7

a + 16d = a + 9d + 7

16d = 9d + 7

16d - 9d = 7

7d = 7

d = 1

Therefore, the common difference of the A.P is 1.

Hope this helps you!!!

Answer 2

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

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

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

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