the n*th term of a sequence is na+b. prove that sequence is an APP. with common difference a.
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Answered by
1
An=na+b
n =1
A1=a+b
A2=2a+b
d=a2-a1
= (2a+b)-(a+b)
=2a-a+b-b
=a
n =1
A1=a+b
A2=2a+b
d=a2-a1
= (2a+b)-(a+b)
=2a-a+b-b
=a
Answered by
1
nth term=na+b
t1=1*a+b
=a+b
t2=2*a+b
=2a+b
t3=3a+b
d1=2a+b-(a+b)
=2a+b-a-b
d =a
d2=3a+b-(2a+b)
=3a+b-2a-b
=a
as common difference is same therefore sequence is on AP
Hence Proved
I hope it will help you!!!!!
t1=1*a+b
=a+b
t2=2*a+b
=2a+b
t3=3a+b
d1=2a+b-(a+b)
=2a+b-a-b
d =a
d2=3a+b-(2a+b)
=3a+b-2a-b
=a
as common difference is same therefore sequence is on AP
Hence Proved
I hope it will help you!!!!!
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