Question

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

Answer 1

Answer:

Step-by-step explanation:

Let the first term be a and common difference be d.

seventeenth term = a + 16d.

tenth term = a + 9d.

NOW, seventeenth term exceeds its tenth term by 7.

∴a + 16d - a - 9d = 7

⇒7d = 7

⇒d = 1

∴Common difference = 1

Answer 2

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

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

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

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