Question

the 17th term of an AP exceeds it's 10th term by 7 . find the common difference.please give step by step​

Answer 1

       

             $\huge\underline{\underline{\bf \bigstar ANSWER \bigstar }}$

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

→  17 th term = a + 16d

→  10 th term = a + 9d

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Given that 17 th term of an AP exceeds its 10 th term by 7 .

$\Longrightarrow \sf 17^{th} \ term - 10^{th } \ term = 7 \\\\\Longrightarrow ( a+16d) - (a+9d) = 7 \\\\\Longrightarrow a+16d-a-9d = 7 \\\\\Longrightarrow 7d = 7 \\\\\Longrightarrow \bf d = 1$______________________________________________

      Common Difference = 1

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Answer 2

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

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

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

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