Question

The seventeenth term of ap exceeds its 10 term by 7. Find the common difference

Answer 1

t17=t10+7

We know: tn=a+(n-1)d
therefore t17=a+(17-1)d =a+16d
similarly. t10=a+(10-1)d =a+9d

SUBSTITUTING THE VALUES

a+16d=a+9d+7
a-a+16d-9d=7
7d=7
d=1

ANSWER common difference(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.

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

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