.
Use the formula for the sum of the first n terms of an arithmetic progression
= = .
Since an arithmetic sequence a1,a2,......a100 has the sum of 15000, you have first equation
= 15000, or
= 300.
The second progression is also an arithmetic progression with the first term of and the common difference of 3d.
Therefore, applying the same formula for the second progression (which has 33 terms), you get
= 5016, or
= 152.
Thus you have this system of 2 equations in 2 unknowns
= 300. (1)
= 152. (2)
Multiply eq(2) by 2 (both sides) and then subtract the eq(1) from it. You will get
d = 304 - 300 = 4.
Then from (2) = 152 - 50*4 = 152 - 200 = - 48.
Answer. = -48; d = 4.
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
- Mathematical induction and arithmetic progressions
- One characteristic property of arithmetic progressions
- Solved problems on arithmetic progressions
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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