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Consider the multiples of 7 in between 100 and 500. a. What are the first and last numbers? b. How many terms are there in the sequence? |
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Answer» a. First term = 100 – 2 + 7 = 105 Last term = 500 – 3 = 497 b. Common difference = xn = dn + (f - d) 497 = 7n + (105 - 7) \(n=\frac{(497-98)}{7}=57\) So, number of terms = 57 |
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