.
The last five terms of this progression are , , , and .
The middle term of these 5 terms is , and it is equal to of the sum of that five terms.
It is the key point in the solution, and you should clearly understand it:
the average of any odd number of sequential terms of an AP is the middle term of this partial sequence.
Thus = = 61.
Now we know = 61 and = 26.
Between and , there are 13-6 = 7 gaps, each of which is equal to the common difference of the AP.
Hence, the common difference d = = = 5.
Then the first term is = - 5*5 = 26 - 25 = 1.
the 15-th term is = + (15-1)*d = 1 + 14*5 = 71.
Now the sum of the 15 terms of the AP is = = 540.
Solved.
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There is a bunch of lessons on arithmetic progressions in this site:
- Arithmetic progressions
- The proofs of the formulas for arithmetic progressions
- Problems on arithmetic progressions
- Word problems on arithmetic progressions
- One characteristic property of arithmetic progressions
- Solved problems on arithmetic progressions
- Calculating partial sums of arithmetic progressions
- Mathematical induction and arithmetic progressions
- Mathematical induction for sequences other than arithmetic or geometric
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".
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Free of charge online textbook in ALGEBRA-II
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