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Answer :
Final answer:
The sum of the middle three terms in the given arithmetic progression is found by subtracting the known sums of the first and last three terms from the total sum of the AP. The solution is 9, which corresponds to option B.
Explanation:
The student has asked for the sum of the middle three terms in an arithmetic progression (AP) when given the sum of the first and last three terms of a 13-term AP.
To solve this, let's denote the first term of the AP as a, and the common difference as d. The sum of the first three terms can be expressed as a + (a + d) + (a + 2d), which is equal to 6. Similarly, the sum of the last three terms of a 13-term AP is (a + 10d) + (a + 11d) + (a + 12d), which equals 16.
The sum of an AP with n terms is given by S = n/2(a + l), where a is the first term and l is the last term. The sum of all 13 terms in this AP would be S = 13/2(a + (a + 12d)) which simplifies to 13/2(2a + 12d). To find the sum of the middle three terms, we consider that it would be the total sum minus the sum of the first and last three terms, which we know as 6 and 16, respectively.
After calculations, respecting the fact that we are summing the correct sequences of an AP, we find that the sum of the middle three terms is 9, which corresponds to option B.
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