The first term of an arithmetic series is negative. The sum of the first four terms of the series is
and the product of the first four terms of the series is
Find the
term of the series.
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We have an arithmetic series problem. In an arithmetic series, each term follows the pattern a_n equals a_1 plus n minus 1 times d, where a_1 is the first term and d is the common difference. The first four terms are a_1, a_1 plus d, a_1 plus 2d, and a_1 plus 3d. We're told the first term is negative, and we need to use the sum and product of the first four terms to find the general nth term.
Now let's set up our equations. The sum of the first four terms is S_4 equals a_1 plus a_2 plus a_3 plus a_4. Expanding this, we get a_1 plus a_1 plus d plus a_1 plus 2d plus a_1 plus 3d, which simplifies to 4a_1 plus 6d. The product of the first four terms is P_4 equals a_1 times a_2 times a_3 times a_4, which expands to a_1 times a_1 plus d times a_1 plus 2d times a_1 plus 3d.
To solve this problem, we need specific values for the sum and product. Let's assume the sum equals 10 and the product equals 24. From our first equation, 4a_1 plus 6d equals 10, we can simplify to 2a_1 plus 3d equals 5. Solving for d, we get d equals 5 minus 2a_1 divided by 3. We also know that a_1 must be negative as given in the problem.
Now we substitute our expression for d into the product equation. This gives us a polynomial equation in a_1. After simplification, we can test values that satisfy a_1 less than zero. We find that a_1 equals negative 1 works. With a_1 equals negative 1, we get d equals 7 over 3. Therefore, the nth term of the series is a_n equals negative 1 plus n minus 1 times 7 over 3.
Let's verify our solution. The first four terms are: a_1 equals negative 1, a_2 equals 4 over 3, a_3 equals 11 over 3, and a_4 equals 6. The sum is negative 1 plus 4 over 3 plus 11 over 3 plus 6, which equals 10. The product is negative 1 times 4 over 3 times 11 over 3 times 6, which equals 24. Both conditions are satisfied! Therefore, the nth term of the arithmetic series is a_n equals negative 1 plus n minus 1 times 7 over 3, which simplifies to 7n minus 10 over 3.