Question

the 11th term of an AP is 60 less than the 31st term of an AP .find the common difference

Answer 1

Answer:

3

Step-by-step explanation:

a11 = a31-60

a+10d = a+30d-60

60= 30d-10d

60=20d

d=3

Answer 2

The common difference is 3.

Given:

The 11th term of an AP is 60 less than the 31st term of an AP.

To Find:

The common difference.

Solution:

To find the common difference we will follow the following steps:

As we know,

The nth term of AP is given by the formula:

$t = a + (n - 1)d$

Here, n is the term number to be found.

a is the first term and d is a common difference.

Now,

11th term is given by:

$t11 = a + (11 - 1)d = a + 10d$

So,

31st term is given by:

$t31 = a + (31 - 1)d = a + 30d$

Now,

The 11th term of an AP is 60 less than the 31st term.

So,

$t31 - t11 = 60$

On putting the value of respective terms we get,

$a + 30d -( a + 10d) = 60$

$a + 30d - a - 10d = 60$

$30d - 10d = 60$

$20d = 60$

$d = \frac{60}{20} = 3$

Henceforth, the common difference is 3.

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