Show that in a quadrilateral ABCD, AB+BC+CD+DA<2(BD+AC)
1 answer:
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You would use the formula for the specific term you wish to find;
The formula is:
a = starting value of the sequence
d = the common difference (i.e. the difference between any two consecutive terms of the sequence)
n = the value corresponding to the position of the desired term in the sequence (i.e. 1 is the first term, 2 is the second, etc.)
Un = the actual vaue of the the term
For example, if we have the arithmetic sequence:
2, 6, 10, 14, ...
And let's say we want to find the 62nd term;
Then:
a = 2
d = 4
(i.e. 6 - 2 = 4, 10 - 6 = 4, 14 - 10 = 4;
You should always get the same number no matter which two terms you find the difference between so long as they are both consecutive [next to each other], otherwise you are not dealing with an arithmetic sequence)
n = 62
And so:
The formula is:
a = starting value of the sequence
d = the common difference (i.e. the difference between any two consecutive terms of the sequence)
n = the value corresponding to the position of the desired term in the sequence (i.e. 1 is the first term, 2 is the second, etc.)
Un = the actual vaue of the the term
For example, if we have the arithmetic sequence:
2, 6, 10, 14, ...
And let's say we want to find the 62nd term;
Then:
a = 2
d = 4
(i.e. 6 - 2 = 4, 10 - 6 = 4, 14 - 10 = 4;
You should always get the same number no matter which two terms you find the difference between so long as they are both consecutive [next to each other], otherwise you are not dealing with an arithmetic sequence)
n = 62
And so:
The answer is 3/5.
Continue this pattern and you'll see that it follows this geometric sequence