in an arithmetic sequence its 3rd term is 64 and 6th term 67 find the sequence.
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Answered by
2
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
Answered by
4
Answer:
294 is the ans it may help you
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