Question 1162164: Find the number of terms in the arithmetic progression 1, 3 , 5,....101
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39623)   (Show Source):
Answer by ikleyn(52835)  (Show Source):
You can put this solution on YOUR website! .
Consider all natural numbers from 1 to 101, inclusive.
Their amount is 101.
From this amount, exclude all even natural numbers from 1 to 100 --- their amount is  = 50.
The difference 101 - 50 = 51 is your answer.
Solved.
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This problem is accessible for the 3-rd - 4-th young students,
who do not know yet neither arithmetic sequences nor equations.
Nevertheless, this problem is a good exercise for the young minds.
So, the solution to the problem must be accessible for them, too.
In other words, the solution should be adequate to the problem.
It is why I came with my post.
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Another path to the answer without algebra, accessible to young students....
The first term is 1; each term is 2 more than the preceding term; the last term is 101.
How many 2's do you need to add to the first term, 1, to get the last term, 101? Answer: (101-1)/2 = 100/2 = 50.
So there are 50 terms after the first one; that means there are 51 terms.
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