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Given that 6,212,...,71 is a linear sequen...

Given that 6,212,...,71 is a linear sequence , calculate the number of terms in the sequence. 

A.

20

B.

21

C.

22

D.

23

Correct answer is D

Tn=a+(n1)d (for a linear or arithmetic progression)

Given: Tn=71,a=6,d=212(6)=312

71=6+(n1)×312

71=6+312n312=912+312n

71+912=312nn=8012312

=23