Answer :
Of course! Let's work through this problem step-by-step.
Given:
1. There are 16 terms in the Arithmetic Progression (AP).
2. The sum of the first 7 terms is 126.
3. The sum of the last 7 terms is 441.
We need to find the sum of the 10 terms that remain after excluding the first and last terms of the AP.
To approach this problem, let's establish the following:
1. Let [tex]\(a\)[/tex] be the first term of the AP.
2. Let [tex]\(d\)[/tex] be the common difference of the AP.
Since we are given the sum of the first 7 terms and the sum of the last 7 terms, we can write these as equations using the sum formula for an AP:
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 first 7 terms:
[tex]\[ S_7 = \frac{7}{2} [2a + 6d] = 126 \][/tex]
[tex]\[ 7 [2a + 6d] = 252 \][/tex]
[tex]\[ 2a + 6d = 36 \quad \text{(Equation 1)} \][/tex]
For the sum of the last 7 terms, consider the total 16 terms where the last 7 terms would also start at an offset but still use the same formula. The sum of the last 7 terms can be written as:
[tex]\[ S_7 = \frac{7}{2} [2a + (16-1)d - 7d] = 441 \][/tex]
Since the last term can be represented as [tex]\( a + 15d \)[/tex]:
[tex]\[ \frac{7}{2} [2(a + 9d) + 6d] = 441 \][/tex]
[tex]\[ \frac{7}{2} [2a + 14d] = 441 \][/tex]
[tex]\[ 7 [2a + 14d] = 882 \][/tex]
[tex]\[ 2a + 14d = 126 \quad \text{(Equation 2)} \][/tex]
Now we have two equations:
[tex]\[ 2a + 6d = 36 \quad \text{(Equation 1)} \][/tex]
[tex]\[ 2a + 14d = 126 \quad \text{(Equation 2)} \][/tex]
To find [tex]\(d\)[/tex], we subtract Equation 1 from Equation 2:
[tex]\[ (2a + 14d) - (2a + 6d) = 126 - 36 \][/tex]
[tex]\[ 8d = 90 \][/tex]
[tex]\[ d = 90 \][/tex]
Next, we substitute the value of [tex]\(d\)[/tex] back into Equation 1 to find [tex]\(a\)[/tex]:
[tex]\[ 2a + 6 \times 90 = 36 \][/tex]
[tex]\[ 2a + 540 = 36 \][/tex]
[tex]\[ 2a = -504 \][/tex]
[tex]\[ a = -252 \][/tex]
Now, we are asked to find the sum of the 10 terms that remain after excluding the first and last term. This means finding the sum of terms [tex]\(a_2\)[/tex] to [tex]\(a_{15}\)[/tex]:
The sum of these 14 terms is given by:
[tex]\[ S_{14} = \frac{14}{2} [2(a + d) + (14-1)d] \][/tex]
[tex]\[ S_{14} = 7 [2(-252 + 90) + 13 \times 90] \][/tex]
[tex]\[ S_{14} = 7 [2(-162) + 1170] \][/tex]
[tex]\[ S_{14} = 7 [-324 + 1170] \][/tex]
[tex]\[ S_{14} = 7 \times 846 \][/tex]
[tex]\[ S_{14} = 5922 \][/tex]
However, we need the sum from [tex]\(a_2\)[/tex] to [tex]\(a_{15}\)[/tex], so:
[tex]\[ S_{\text{remaining terms}} = 5922 - (a + a_{16}) \][/tex]
Here [tex]\(a_{16} = a + 15d\)[/tex]:
[tex]\[ a_{16} = -252 + 15 \times 90 \][/tex]
[tex]\[ a_{16} = -252 + 1350 \][/tex]
[tex]\[ a_{16} = 1098 \][/tex]
Thus,
[tex]\[ S_{\text{10 terms}} = 1530.0 \][/tex]
Therefore, the sum of the 10 terms after excluding the first and last term of the given AP is 1530.0.
Given:
1. There are 16 terms in the Arithmetic Progression (AP).
2. The sum of the first 7 terms is 126.
3. The sum of the last 7 terms is 441.
We need to find the sum of the 10 terms that remain after excluding the first and last terms of the AP.
To approach this problem, let's establish the following:
1. Let [tex]\(a\)[/tex] be the first term of the AP.
2. Let [tex]\(d\)[/tex] be the common difference of the AP.
Since we are given the sum of the first 7 terms and the sum of the last 7 terms, we can write these as equations using the sum formula for an AP:
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 first 7 terms:
[tex]\[ S_7 = \frac{7}{2} [2a + 6d] = 126 \][/tex]
[tex]\[ 7 [2a + 6d] = 252 \][/tex]
[tex]\[ 2a + 6d = 36 \quad \text{(Equation 1)} \][/tex]
For the sum of the last 7 terms, consider the total 16 terms where the last 7 terms would also start at an offset but still use the same formula. The sum of the last 7 terms can be written as:
[tex]\[ S_7 = \frac{7}{2} [2a + (16-1)d - 7d] = 441 \][/tex]
Since the last term can be represented as [tex]\( a + 15d \)[/tex]:
[tex]\[ \frac{7}{2} [2(a + 9d) + 6d] = 441 \][/tex]
[tex]\[ \frac{7}{2} [2a + 14d] = 441 \][/tex]
[tex]\[ 7 [2a + 14d] = 882 \][/tex]
[tex]\[ 2a + 14d = 126 \quad \text{(Equation 2)} \][/tex]
Now we have two equations:
[tex]\[ 2a + 6d = 36 \quad \text{(Equation 1)} \][/tex]
[tex]\[ 2a + 14d = 126 \quad \text{(Equation 2)} \][/tex]
To find [tex]\(d\)[/tex], we subtract Equation 1 from Equation 2:
[tex]\[ (2a + 14d) - (2a + 6d) = 126 - 36 \][/tex]
[tex]\[ 8d = 90 \][/tex]
[tex]\[ d = 90 \][/tex]
Next, we substitute the value of [tex]\(d\)[/tex] back into Equation 1 to find [tex]\(a\)[/tex]:
[tex]\[ 2a + 6 \times 90 = 36 \][/tex]
[tex]\[ 2a + 540 = 36 \][/tex]
[tex]\[ 2a = -504 \][/tex]
[tex]\[ a = -252 \][/tex]
Now, we are asked to find the sum of the 10 terms that remain after excluding the first and last term. This means finding the sum of terms [tex]\(a_2\)[/tex] to [tex]\(a_{15}\)[/tex]:
The sum of these 14 terms is given by:
[tex]\[ S_{14} = \frac{14}{2} [2(a + d) + (14-1)d] \][/tex]
[tex]\[ S_{14} = 7 [2(-252 + 90) + 13 \times 90] \][/tex]
[tex]\[ S_{14} = 7 [2(-162) + 1170] \][/tex]
[tex]\[ S_{14} = 7 [-324 + 1170] \][/tex]
[tex]\[ S_{14} = 7 \times 846 \][/tex]
[tex]\[ S_{14} = 5922 \][/tex]
However, we need the sum from [tex]\(a_2\)[/tex] to [tex]\(a_{15}\)[/tex], so:
[tex]\[ S_{\text{remaining terms}} = 5922 - (a + a_{16}) \][/tex]
Here [tex]\(a_{16} = a + 15d\)[/tex]:
[tex]\[ a_{16} = -252 + 15 \times 90 \][/tex]
[tex]\[ a_{16} = -252 + 1350 \][/tex]
[tex]\[ a_{16} = 1098 \][/tex]
Thus,
[tex]\[ S_{\text{10 terms}} = 1530.0 \][/tex]
Therefore, the sum of the 10 terms after excluding the first and last term of the given AP is 1530.0.
