Question

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

Answer 1

hey mate here your answer
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d=?

t17-t10=7

a+16d-a-9d=7

a-a+16d-9d=7


7d=7

d=1

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thanks mate

Answer 2

Hey there!

Arithmetic progression : It is a series in which the difference between two consecutive terms is constant.

General form of n-th term,
$a_{n} = a + ( n - 1 ) d$
So,
Let a be the first term, and common difference be d.

17th term = a + ( 17 - 1 ) d = a + 16d

10th term = a + ( 10 - 1 ) d = a + 9d .

Given that,

$\textbf{ The 17th term of an AP exceeds its 10th term by 7}$

$\mathbf { a_{17} - 7 = a_{10} }$

a + 16d - 7 = a + 9d

16d - 7 = 9d

16d - 9d = 7

7d = 7

d = 1 .

Therefore, The common difference is 1