To solve the problem, we need to find the 10th term of an arithmetic progression (AP) given that the 6th term is 56 and the sum of the first four terms is 70.
### Step 1: Represent the given information in terms of the first term [tex]\(a\)[/tex] and the common difference [tex]\(d\)[/tex].
1. The [tex]\(n\)[/tex]-th term of an AP is given by:
[tex]\[ a_n = a + (n-1)d \][/tex]
For the 6th term:
[tex]\[ a + 5d = 56 \ \ \ \ \ \ \ \ \ \ \ (1) \][/tex]
2. The sum of the first [tex]\(n\)[/tex] terms of an AP is given by:
[tex]\[ S_n = \frac{n}{2} [2a + (n-1)d] \][/tex]
For the sum of the first 4 terms:
[tex]\[ S_4 = \frac{4}{2} [2a + 3d] \][/tex]
[tex]\[ S_4 = 2 (2a + 3d) \][/tex]
[tex]\[ 4a + 6d = 70 \ \ \ \ \ \ \ \ \ \ \ (2) \][/tex]
### Step 2: Solve the system of linear equations to find [tex]\(a\)[/tex] and [tex]\(d\)[/tex].
We now have the following system of equations:
1. [tex]\( a + 5d = 56 \)[/tex]
2. [tex]\( 4a + 6d = 70 \)[/tex]
Let's solve these equations step-by-step.
#### From Equation (1):
[tex]\[ a = 56 - 5d \ \ \ \ \ \ \ \ \ \ \ (3) \][/tex]
#### Substitute equation (3) into Equation (2):
[tex]\[ 4(56 - 5d) + 6d = 70 \][/tex]
[tex]\[ 224 - 20d + 6d = 70 \][/tex]
[tex]\[ 224 - 14d = 70 \][/tex]
[tex]\[ -14d = 70 - 224 \][/tex]
[tex]\[ -14d = -154 \][/tex]
[tex]\[ d = \frac{-154}{-14} \][/tex]
[tex]\[ d = 11 \][/tex]
Now substitute [tex]\(d = 11\)[/tex] back into Equation (3):
[tex]\[ a = 56 - 5 \cdot 11 \][/tex]
[tex]\[ a = 56 - 55 \][/tex]
[tex]\[ a = 1 \][/tex]
### Step 3: Calculate the 10th term of the AP.
The 10th term of an AP is given by:
[tex]\[ a_{10} = a + 9d \][/tex]
Substituting [tex]\(a = 1\)[/tex] and [tex]\(d = 11\)[/tex]:
[tex]\[ a_{10} = 1 + 9 \cdot 11 \][/tex]
[tex]\[ a_{10} = 1 + 99 \][/tex]
[tex]\[ a_{10} = 100 \][/tex]
Therefore, the 10th term of the AP is [tex]\( \boxed{100} \)[/tex].
