What is a1 of the arithmetic sequence for which a3=126 and a64= 3,725
1 answer:
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.
Learn more about arithmetic progression at brainly.com/question/6561461
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