Question

The 3rd term of an arithmetic sequence is 32 and the 8th term is 164. what is the first term?

Answer 1

Step-by-step explanation:

ANSWER

t3=34 and t8=69

(a) Now, tn=a+(n−1)d

⇒t3=a+(2−1)d and t8=a+(7−1)d

⇒34=a+(2−1)d and 69=a+(7−1)d

⇒34=a+d and 69=a+6d

Thus, 69−34=a+6d−a−d

⇒35=5d⇒d=7

(b) First term, a=t3−2d=34−2(7)=34−14=20

General term, tn=a+(n−1)d=20+(n−1)7=20+7n−7=7n+13

Thus, tn=7n+13 is the algebraic form of this sequence.

(c) tn=7n+13

If each term of the sequence is multiplied by 4 and then 3 is added to it, the

general term will be tn=[(7n+13)×