Certainly! Let's begin by identifying the arithmetic progression (AP):
Given:
1. The sum of the first four terms is 46.
2. The seventh term exceeds twice the second term by 5.
### Step 1: Set up the equations
Let the first term be [tex]\( a \)[/tex] and the common difference be [tex]\( d \)[/tex].
The first four terms of the AP are:
[tex]\[ a, a + d, a + 2d, a + 3d \][/tex]
The sum of these terms is:
[tex]\[ a + (a + d) + (a + 2d) + (a + 3d) = 46 \][/tex]
[tex]\[ 4a + 6d = 46 \][/tex]
[tex]\[ 2a + 3d = 23 \][/tex]
This is our first equation.
The seventh term of the AP is:
[tex]\[ a + 6d \][/tex]
The second term of the AP is:
[tex]\[ a + d \][/tex]
Twice the second term is:
[tex]\[ 2(a + d) = 2a + 2d \][/tex]
Given that the seventh term exceeds twice the second term by 5:
[tex]\[ a + 6d = 2a + 2d + 5 \][/tex]
Simplify this to get:
[tex]\[ a + 6d = 2a + 2d + 5 \][/tex]
[tex]\[ a + 6d - 2a - 2d = 5 \][/tex]
[tex]\[ -a + 4d = 5 \][/tex]
[tex]\[ a = 4d - 5 \][/tex]
This is our second equation.
### Step 2: Solve the system of equations
We have two equations:
1. [tex]\( 2a + 3d = 23 \)[/tex]
2. [tex]\( a = 4d - 5 \)[/tex]
Substitute [tex]\( a = 4d - 5 \)[/tex] into the first equation:
[tex]\[ 2(4d - 5) + 3d = 23 \][/tex]
[tex]\[ 8d - 10 + 3d = 23 \][/tex]
[tex]\[ 11d - 10 = 23 \][/tex]
[tex]\[ 11d = 33 \][/tex]
[tex]\[ d = 3 \][/tex]
Now, substitute [tex]\( d = 3 \)[/tex] back into [tex]\( a = 4d - 5 \)[/tex]:
[tex]\[ a = 4(3) - 5 \][/tex]
[tex]\[ a = 12 - 5 \][/tex]
[tex]\[ a = 7 \][/tex]
Thus, the first term [tex]\( a \)[/tex] is 7 and the common difference [tex]\( d \)[/tex] is 3.
### Step 3: Sum of the first ten even terms
Even terms in an AP sequence are also in arithmetic progression, with each term being even. The first, second, third, etc., even terms can be expressed as follows:
- The first even term will be [tex]\( 2(a + nd) \)[/tex], where [tex]\( n \)[/tex] is 0, 1, 2, etc.
The first ten even terms of the sequence are:
[tex]\[ 2a, 2(a + d), 2(a + 2d), \ldots, 2(a + 9d) \][/tex]
This is also an arithmetic progression with the first term [tex]\( 2a \)[/tex] and common difference [tex]\( 2d \)[/tex]:
[tex]\[ 2a, 2a + 2d, 2a + 4d, \ldots, 2a + 18d \][/tex]
The sum [tex]\( S \)[/tex] of the first [tex]\( n \)[/tex] terms of an AP is given by:
[tex]\[ S_n = \frac{n}{2} [2a' + (n-1)d'] \][/tex]
where [tex]\( n = 10 \)[/tex], [tex]\( a' = 2a \)[/tex], and [tex]\( d' = 2d \)[/tex].
Substituting [tex]\( a = 7 \)[/tex] and [tex]\( d = 3 \)[/tex], we get:
[tex]\[ a' = 2a = 2(7) = 14 \][/tex]
[tex]\[ d' = 2d = 2(3) = 6 \][/tex]
So, the sum of the first ten even terms:
[tex]\[ S_{10} = \frac{10}{2} [2(14) + (10-1)(6)] \][/tex]
[tex]\[ S_{10} = 5 [28 + 54] \][/tex]
[tex]\[ S_{10} = 5 \times 82 \][/tex]
[tex]\[ S_{10} = 410 \][/tex]
Therefore, the sum of the first ten even terms of the progression is 410.
