if the sum of the 6th and 7th term of an Arithmetic progression(AP) is 60 and the 3rd term is -5 find the common difference?

To find the common difference of an arithmetic progression (AP) when given specific terms, we can use the formula:

nth term = a + (n - 1) * d

where:
- nth term is the term number we want to find
- a is the first term of the AP
- n is the term number we want to find
- d is the common difference of the AP

Given that the 3rd term of the AP is -5, we can write it as:

-5 = a + (3 - 1) * d

Simplifying the equation, we get:

-5 = a + 2d

Now, we are also given that the sum of the 6th and 7th terms is 60, which gives us:

6th term + 7th term = 60

Using the same formula, we can write it as:

a + (6 - 1) * d + a + (7 - 1) * d = 60

Simplifying the equation, we get:

2a + 12d = 60

Now, we have two equations:

-5 = a + 2d ...(1)
2a + 12d = 60 ...(2)

We can solve these two equations to find the common difference (d).

One approach is to solve equation (1) for "a" and substitute it into equation (2).

From equation (1), we get:

a = -5 - 2d

Substituting this into equation (2), we have:

2(-5 - 2d) + 12d = 60

Simplifying the equation, we get:

-10 - 4d + 12d = 60

Combining like terms, we get:

8d - 10 = 60

Adding 10 to both sides of the equation, we get:

8d = 70

Dividing both sides of the equation by 8, we get:

d = 8.75

Therefore, the common difference of the arithmetic progression is 8.75.

To find the common difference of an arithmetic progression (AP) given certain terms and their values, you can use the formula:

\[a_n = a_1 + (n - 1)d\]

Where:
\(a_n\) is the \(n\)th term of the AP,
\(a_1\) is the first term of the AP,
\(n\) is the position of the term,
\(d\) is the common difference.

Let's use this formula to solve the problem.

Given:
The 3rd term, \(a_3\) = -5
The sum of the 6th and 7th terms, \(a_6 + a_7\) = 60

First, we can find the values of the 6th and 7th terms by substituting their positions into the formula.

For the 6th term (\(a_6\)):
\(a_6 = a_1 + (6-1)d = a_1 + 5d\)

For the 7th term (\(a_7\)):
\(a_7 = a_1 + (7-1)d = a_1 + 6d\)

We are given that \(a_6 + a_7 = 60\). Substituting the expressions for \(a_6\) and \(a_7\) into the equation, we get:

\(a_1 + 5d + a_1 + 6d = 60\)

Simplifying, we have:

\(2a_1 + 11d = 60\)

Now, we are also given that \(a_3 = -5\). Substituting \(a_3\) into the formula, we get:

\(a_3 = a_1 + (3-1)d\)

Simplifying, we have:

\(a_1 + 2d = -5\)

We now have a system of two equations:
\(2a_1 + 11d = 60\)
\(a_1 + 2d = -5\)

To solve this system, you can use any method you prefer, such as substitution or elimination. Let's use the elimination method here.

Multiply the second equation by 2 to make the coefficients of \(a_1\) match:

\(2(a_1 + 2d) = -10\)

Simplifying, we get:

\(2a_1 + 4d = -10\)

Now, subtract the new equation from the first equation to eliminate \(a_1\):

\((2a_1 + 11d) - (2a_1 + 4d) = 60 - (-10)\)

Simplifying, we have:

\(7d = 70\)

Divide both sides by 7:

\(d = 10\)

Therefore, the common difference of the arithmetic progression (AP) is 10.

6th is ... -5 + 3d

7th is ... -5 + 4d

-10 + 7d = 60