1 +4 +8 +9 +12 +16+ 20 + 24 +27+ 28....729 +732. STATE THE FIRST THREE TERMS AND THE. LAST TERM OF THE ARITHMETIC SERIES
the series is the sequence of partial sums. So, the first three terms are
1,5,13
No idea what the last term is, since I don't see a pattern to the items.
Write the first ten terms of a sequence whose first term is -10 and whose common difference is -2.
To find the first three terms and the last term of the arithmetic series, we need to understand the pattern of the series.
In an arithmetic series, each term is obtained by adding a constant difference to the previous term.
The first term (a) can be determined from the given series, which is 1.
The common difference (d) can be calculated by subtracting the second term from the first term. In this case, it is 4 - 1 = 3.
With these values, we can find the first three terms of the series:
First term (a) = 1
Second term = first term + common difference = 1 + 3 = 4
Third term = second term + common difference = 4 + 3 = 7
So, the first three terms of the series are 1, 4, and 7.
To find the last term, we need to determine the number of terms (n) in the series. You can do this by finding the difference between the last term (732) and the first term (1), and then dividing it by the common difference (3).
Number of terms (n) = (last term - first term) / common difference
= (732 - 1) / 3
= 731 / 3
However, this will give us a decimal value, indicating that the series is not complete. It means that 732 is not the last term of the series. We need to find the closest number divisible by 3 that is greater than 732.
Dividing 732 by 3, we get 244 with a remainder.
So, the next multiple of 3 after 732 is 735.
Hence, the last term of the series is 735.
In summary,
First term: 1
Second term: 4
Third term: 7
Last term: 735