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Answer :
So 1.70 (you can multiply it by 100 to make things easier if you want) - 1.50 = 0.15. So 0.15 is the difference between one term and the next.
Also, if you subtract 1.40 from 1.70 you'll find that the difference is 0.30, which is twice 0.15. And 1.70 is two terms away from 1.40. Thus, 1.40 + 0.15m is true, (m being the amount of terms away from the starting number.) In the case I mentioned, 1.40 + 0.15(2) was true. So, plug 31 into the equation:
1.40 + (31)(0.15) = 6.05. This is your answer.
Also, if you subtract 1.40 from 1.70 you'll find that the difference is 0.30, which is twice 0.15. And 1.70 is two terms away from 1.40. Thus, 1.40 + 0.15m is true, (m being the amount of terms away from the starting number.) In the case I mentioned, 1.40 + 0.15(2) was true. So, plug 31 into the equation:
1.40 + (31)(0.15) = 6.05. This is your answer.
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Rewritten by : Barada
To find the 31st term of the arithmetic sequence with a common difference of 0.15, we apply the formula for the nth term of an arithmetic sequence. The 31st term is calculated as 1.40 plus 30 times the common difference, resulting in a value of 5.90.
To find the 31st term of the given sequence, we need to determine the pattern or formula that generates the terms of the sequence. The numbers provided, 1.40, 1.55, and 1.70, suggest that this is an arithmetic sequence because the difference between consecutive terms is constant.
This constant difference is also known as the common difference. In this case, the common difference is 1.70 - 1.55 = 0.15. Knowing this, we can generate the 31st term using the arithmetic sequence formula:
nth term = first term + (n - 1) × common difference
Substituting the known values into the formula, we have:
31st term = 1.40 + (31 - 1) × 0.15
31st term = 1.40 + 30 × 0.15
31st term = 1.40 + 4.50
31st term = 5.90
Therefore, the 31st term of the sequence is 5.90.
