Sure, let's solve this problem step by step.
Given:
1. The sum of the 6th and 7th terms of an arithmetic progression (AP) is 60.
2. The 3rd term of the AP is -5.
We can use the following notations:
- Let [tex]\( a \)[/tex] be the first term of the arithmetic progression.
- Let [tex]\( d \)[/tex] be the common difference of the progression.
The general formula for the [tex]\( n \)[/tex]-th term of an AP is given by:
[tex]\[ a_n = a + (n-1)d \][/tex]
### Step 1: Determine the expressions for the relevant terms.
- The 3rd term of the AP [tex]\( a_3 \)[/tex] can be written as:
[tex]\[ a_3 = a + 2d \][/tex]
- The 6th term of the AP [tex]\( a_6 \)[/tex] can be written as:
[tex]\[ a_6 = a + 5d \][/tex]
- The 7th term of the AP [tex]\( a_7 \)[/tex] can be written as:
[tex]\[ a_7 = a + 6d \][/tex]
### Step 2: Use the given information.
- From the given information, we know:
[tex]\[ a + 5d + a + 6d = 60 \][/tex]
This simplifies to:
[tex]\[ 2a + 11d = 60 \][/tex]
- We are also given that the 3rd term is -5:
[tex]\[ a + 2d = -5 \][/tex]
### Step 3: Solve the system of equations.
We have two equations now:
1. [tex]\( 2a + 11d = 60 \)[/tex]
2. [tex]\( a + 2d = -5 \)[/tex]
Let's solve these equations simultaneously.
Equation 1:
[tex]\[ 2a + 11d = 60 \][/tex]
Equation 2:
[tex]\[ a + 2d = -5 \][/tex]
We can multiply Equation 2 by 2 to facilitate elimination:
[tex]\[ 2(a + 2d) = 2(-5) \][/tex]
[tex]\[ 2a + 4d = -10 \][/tex]
Now we have:
1. [tex]\( 2a + 11d = 60 \)[/tex]
2. [tex]\( 2a + 4d = -10 \)[/tex]
Subtract the second equation from the first equation to eliminate [tex]\( a \)[/tex]:
[tex]\[ (2a + 11d) - (2a + 4d) = 60 - (-10) \][/tex]
[tex]\[ 2a + 11d - 2a - 4d = 60 + 10 \][/tex]
[tex]\[ 7d = 70 \][/tex]
### Step 4: Solve for [tex]\( d \)[/tex].
[tex]\[ d = \frac{70}{7} \][/tex]
[tex]\[ d = 10 \][/tex]
### Conclusion:
The common difference [tex]\( d \)[/tex] of the arithmetic progression is [tex]\( 10 \)[/tex].
