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MIP1502 assignment 3 due 20 JULY 2023

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Mathematics for Intermediate Phase Teachers II MIP1502 Year module Department of Mathematics Education ASSESSMENT 3 UNIQUE NO: DUE ATE : 04 JULY 2023 Question 1 1.1 If Lize would continue counting 3; 8; 13; …. 1.1.1 What is the 100th number that she will count? Show Lize how to calculate the 100th number instead of actually counting all the way up to the 100th number. (6) 1.1.1 To find the 100th number that Lize will count, we can use the formula for an arithmetic sequence. The general formula for the nth term of an arithmetic sequence is: a_n = a_1 + (n - 1) * d where a_n is the nth term, a_1 is the first term, n is the position number, and d is the common difference between terms. In this case, the first term (a_1) is 3, and the common difference (d) is 5 (since each number is 5 more than the previous one). Using the formula, we can substitute the values: a_100 = 3 + (100 - 1) * 5 = 3 + 99 * 5 = 3 + 495 = 498 Therefore, the 100th number that Lize will count is 498.

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Mathematics for Intermediate Phase Teachers II

MIP1502

Year module

Department of Mathematics Education

ASSESSMENT 3

UNIQUE NO: 787060


DUE ATE : 04 JULY 2023

, Question 1




1.1 If Lize would continue counting 3; 8; 13; ….




1.1.1 What is the 100th number that she will count? Show Lize how to calculate the

100th number instead of actually counting all the way up to the 100th number. (6)

1.1.1 To find the 100th number that Lize will count, we can use the formula for an

arithmetic sequence. The general formula for the nth term of an arithmetic sequence is:

a_n = a_1 + (n - 1) * d

where a_n is the nth term, a_1 is the first term, n is the position number, and d is the

common difference between terms.

In this case, the first term (a_1) is 3, and the common difference (d) is 5 (since each

number is 5 more than the previous one).

Using the formula, we can substitute the values:

a_100 = 3 + (100 - 1) * 5

= 3 + 99 * 5

= 3 + 495

= 498

Therefore, the 100th number that Lize will count is 498.

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