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To find the value of [tex]\( n \)[/tex] for which the sum of the first [tex]\( n \)[/tex] terms of the arithmetic progression (A.P.) [tex]\( -4, 0, 4, 8, \ldots \)[/tex] is equal to the sum of the first [tex]\( n \)[/tex] terms of the arithmetic progression (A.P.) [tex]\( 2, 4, 6, 8, \ldots \)[/tex], let's proceed step-by-step.
### Step 1: Definitions and Formulas
The general formula for the sum of the first [tex]\( n \)[/tex] terms ([tex]\( S_n \)[/tex]) of an arithmetic progression (A.P.) is:
[tex]\[ S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right] \][/tex]
where [tex]\( a \)[/tex] is the first term and [tex]\( d \)[/tex] is the common difference.
### Step 2: Identify Parameters for Each A.P.
First A.P.: [tex]\( -4, 0, 4, 8, \ldots \)[/tex]
- First term ([tex]\( a_1 \)[/tex]) = [tex]\( -4 \)[/tex]
- Common difference ([tex]\( d_1 \)[/tex]) = [tex]\( 0 - (-4) = 4 \)[/tex]
Second A.P.: [tex]\( 2, 4, 6, 8, \ldots \)[/tex]
- First term ([tex]\( a_2 \)[/tex]) = [tex]\( 2 \)[/tex]
- Common difference ([tex]\( d_2 \)[/tex]) = [tex]\( 4 - 2 = 2 \)[/tex]
### Step 3: Sum of the First [tex]\( n \)[/tex] Terms for Each A.P.
Sum of the first [tex]\( n \)[/tex] terms of the first A.P.:
[tex]\[ S_{1,n} = \frac{n}{2} \left[ 2(-4) + (n - 1) \cdot 4 \right] \][/tex]
[tex]\[ S_{1,n} = \frac{n}{2} \left[ -8 + 4n - 4 \right] \][/tex]
[tex]\[ S_{1,n} = \frac{n}{2} \left[ 4n - 12 \right] \][/tex]
[tex]\[ S_{1,n} = \frac{n}{2} \cdot 4(n - 3) \][/tex]
[tex]\[ S_{1,n} = 2n(n - 3) \][/tex]
[tex]\[ S_{1,n} = 2n^2 - 6n \][/tex]
Sum of the first [tex]\( n \)[/tex] terms of the second A.P.:
[tex]\[ S_{2,n} = \frac{n}{2} \left[ 2(2) + (n - 1) \cdot 2 \right] \][/tex]
[tex]\[ S_{2,n} = \frac{n}{2} \left[ 4 + 2n - 2 \right] \][/tex]
[tex]\[ S_{2,n} = \frac{n}{2} \left[ 2n + 2 \right] \][/tex]
[tex]\[ S_{2,n} = \frac{n}{2} \cdot 2(n + 1) \][/tex]
[tex]\[ S_{2,n} = n(n + 1) \][/tex]
[tex]\[ S_{2,n} = n^2 + n \][/tex]
### Step 4: Setting the Sums Equal and Solving for [tex]\( n \)[/tex]
Set [tex]\( S_{1,n} \)[/tex] equal to [tex]\( S_{2,n} \)[/tex]:
[tex]\[ 2n^2 - 6n = n^2 + n \][/tex]
Move all terms to one side to form a quadratic equation:
[tex]\[ 2n^2 - 6n - n^2 - n = 0 \][/tex]
[tex]\[ n^2 - 7n = 0 \][/tex]
Factor out [tex]\( n \)[/tex]:
[tex]\[ n(n - 7) = 0 \][/tex]
Set each factor to zero:
[tex]\[ n = 0 \quad \text{or} \quad n - 7 = 0 \][/tex]
[tex]\[ n = 0 \quad \text{or} \quad n = 7 \][/tex]
### Step 5: Conclusion
The values of [tex]\( n \)[/tex] for which the sums of the first terms of both arithmetic progressions are equal are:
[tex]\[ n = 0 \quad \text{or} \quad n = 7 \][/tex]
Since [tex]\( n = 0 \)[/tex] does not make sense in the context of summing the first [tex]\( n \)[/tex] terms, we discard [tex]\( n = 0 \)[/tex].
Thus, the value of [tex]\( n \)[/tex] is:
[tex]\[ n = 7 \][/tex]
### Step 1: Definitions and Formulas
The general formula for the sum of the first [tex]\( n \)[/tex] terms ([tex]\( S_n \)[/tex]) of an arithmetic progression (A.P.) is:
[tex]\[ S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right] \][/tex]
where [tex]\( a \)[/tex] is the first term and [tex]\( d \)[/tex] is the common difference.
### Step 2: Identify Parameters for Each A.P.
First A.P.: [tex]\( -4, 0, 4, 8, \ldots \)[/tex]
- First term ([tex]\( a_1 \)[/tex]) = [tex]\( -4 \)[/tex]
- Common difference ([tex]\( d_1 \)[/tex]) = [tex]\( 0 - (-4) = 4 \)[/tex]
Second A.P.: [tex]\( 2, 4, 6, 8, \ldots \)[/tex]
- First term ([tex]\( a_2 \)[/tex]) = [tex]\( 2 \)[/tex]
- Common difference ([tex]\( d_2 \)[/tex]) = [tex]\( 4 - 2 = 2 \)[/tex]
### Step 3: Sum of the First [tex]\( n \)[/tex] Terms for Each A.P.
Sum of the first [tex]\( n \)[/tex] terms of the first A.P.:
[tex]\[ S_{1,n} = \frac{n}{2} \left[ 2(-4) + (n - 1) \cdot 4 \right] \][/tex]
[tex]\[ S_{1,n} = \frac{n}{2} \left[ -8 + 4n - 4 \right] \][/tex]
[tex]\[ S_{1,n} = \frac{n}{2} \left[ 4n - 12 \right] \][/tex]
[tex]\[ S_{1,n} = \frac{n}{2} \cdot 4(n - 3) \][/tex]
[tex]\[ S_{1,n} = 2n(n - 3) \][/tex]
[tex]\[ S_{1,n} = 2n^2 - 6n \][/tex]
Sum of the first [tex]\( n \)[/tex] terms of the second A.P.:
[tex]\[ S_{2,n} = \frac{n}{2} \left[ 2(2) + (n - 1) \cdot 2 \right] \][/tex]
[tex]\[ S_{2,n} = \frac{n}{2} \left[ 4 + 2n - 2 \right] \][/tex]
[tex]\[ S_{2,n} = \frac{n}{2} \left[ 2n + 2 \right] \][/tex]
[tex]\[ S_{2,n} = \frac{n}{2} \cdot 2(n + 1) \][/tex]
[tex]\[ S_{2,n} = n(n + 1) \][/tex]
[tex]\[ S_{2,n} = n^2 + n \][/tex]
### Step 4: Setting the Sums Equal and Solving for [tex]\( n \)[/tex]
Set [tex]\( S_{1,n} \)[/tex] equal to [tex]\( S_{2,n} \)[/tex]:
[tex]\[ 2n^2 - 6n = n^2 + n \][/tex]
Move all terms to one side to form a quadratic equation:
[tex]\[ 2n^2 - 6n - n^2 - n = 0 \][/tex]
[tex]\[ n^2 - 7n = 0 \][/tex]
Factor out [tex]\( n \)[/tex]:
[tex]\[ n(n - 7) = 0 \][/tex]
Set each factor to zero:
[tex]\[ n = 0 \quad \text{or} \quad n - 7 = 0 \][/tex]
[tex]\[ n = 0 \quad \text{or} \quad n = 7 \][/tex]
### Step 5: Conclusion
The values of [tex]\( n \)[/tex] for which the sums of the first terms of both arithmetic progressions are equal are:
[tex]\[ n = 0 \quad \text{or} \quad n = 7 \][/tex]
Since [tex]\( n = 0 \)[/tex] does not make sense in the context of summing the first [tex]\( n \)[/tex] terms, we discard [tex]\( n = 0 \)[/tex].
Thus, the value of [tex]\( n \)[/tex] is:
[tex]\[ n = 7 \][/tex]
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