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The sum of the first 30 terms in an arithmetic sequence is 1830 and the 8th term is 31 determine the first three terms​

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User Shadox
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Answer:

Let's use the formula for the sum of the first n terms of an arithmetic sequence to find the value of the common difference d:

S = n/2 * (2a1 + (n-1)d), where S is the sum of the first n terms, a1 is the first term, and d is the common difference.

We know that S = 1830 and n = 30, so we can write:

1830 = 30/2 * (2a1 + 29d)

1830 = 15(2a1 + 29d)

122 = 2a1 + 29d

Next, we know that the 8th term is 31, so we can write:

a8 = a1 + 7d = 31

Now we have two equations with two unknowns, so we can solve for a1 and d:

122 = 2a1 + 29d

31 = a1 + 7d

Multiplying the second equation by 2 and subtracting it from the first equation, we get:

60 = 15d

So d = 4.

Substituting d = 4 into the equation a1 + 7d = 31, we get:

a1 + 28 = 31

a1 = 3

Therefore, the first three terms of the arithmetic sequence are:

a1 = 3

a2 = a1 + d = 3 + 4 = 7

a3 = a2 + d = 7 + 4 = 11

So the first three terms are 3, 7, 11.

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User Huonderv
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