If the third term of the aritmetic sequence is 126 and sixty fourth term is 3725 then the first term is 8.
Given the third term of the aritmetic sequence is 126 and sixty fourth term is 3725.
We are required to find the first term of the arithmetic sequence.
Arithmetic sequence is a series in which all the terms have equal difference.
Nth term of an AP=a+(n-1)d
=a+(3-1)d
126=a+2d--------1
=a+(64-1)d
3725=a+63d------2
Subtract second equation from first equation.
a+2d-a-63d=126-3725
-61d=-3599
d=59
Put the value of d in 1 to get the value of a.
a+2d=126
a+2*59=126
a+118=126
a=126-118
a=8
=a+(1-1)d
=8+0*59
=8
Hence if the third term of the arithmetic sequence is 126 and sixty fourth term is 3725 then the first term is 8.
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(3 - 0.49) / 3 = 2.51/3 = 0.836 = 83.6% is saved
Answer:

Step-by-step explanation:
Using the rule of exponents
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Answer:
Step-by-step explanation:
sin(32°) = b/340
b = 340sin(32°) ≅ 180 inches
Answer:
i believe its x= 16
Step-by-step explanation:
(4/2) 2x/4 = 8/1 (4/2)
after you do that it should be
x = 32/2, now just simplify
x=16