Question

in an arithmetic sequence its 3rd term is 64 and 6th term 67 find the sequence. ​

Answer 1

Answer:

t

3

=34 and t

8

=69

(a) Now, t

n

=a+(n−1)d

⇒t

3

=a+(2−1)d and t

8

=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=t

3

−2d=34−2(7)=34−14=20

General term, t

n

=a+(n−1)d=20+(n−1)7=20+7n−7=7n+13

Thus, t

n

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

(c) t

n

=7n+13

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

general term will be t

n

=[(7n+13)×4]+3=28n+52+3=28n+55

Thus, tenth term, t

10

=28(10)+55=280+55=335

Answer 2

Answer:

294 is the ans it may help you