Question

If the 17th term of an AP exceeds its 10th term by 7, find the common difference. ​

Answer 1

Answer:

Let the first term be a.

The common difference is d.

According to the question ;

$t17 = a + (17 - 1)d$

$t10 = a + (10 - 1)d$

$As \: the \: 17th \: term \: exceeds \: the \: 10th \: term \: by \: 7 \: so$

$= > (a + 16d) - (a + 9d) = 7$

$= > 7d \: = 7$

$= > d \: = \: 1$

Answer 2

Answer:

D = 1

Explanation:

a17 = a+16d

a10 = a+9d

given that,

a+16d =a+ 9d +7

16d = 9d + 7

16 d - 9d =7

7d =7

d =1