Find the 11th term from the end in the AP 56, 63, 70,..., 329

1 answer:

4 0

Given sequence:
56, 63, 70,..., 329
First find the common difference:
d=a2-a1
d=63-56
d=7
Formula:
an=a1+(n-1)d
First find the number of terms.
329=56+(n-1)7
329=56+7n-7
329=49+7n
280=7n
40=n
or
n=40
We have to find 40-11=29 th term ( 11 th term from the end)
a29= 56+(29-1)7
a29=56+(28)7
a29=56+196
a29=252

Answer: 252 is 11th term from the end.

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<h3>Pythagorean relation</h3>

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This can be solved for sin(x) in terms of cos(x):

sin(x) = √(1 -cos(x)²)

<h3>Application</h3>

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<em>Additional comment</em>

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