Q.5: Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?
Answers
Answered by
5
Step-by-step explanation:
its 54th is
3 + (54- 1) × 12
3+ (53) ×12
639
then n term is qual to 54th term + 132
771 = 3+(n-1)×12
771 = 3+ 12n -12
771-3+12 =12 n
800 = 12n something went wrong , try this way for right answer.
Answered by
11
☆ANSWER☆
Step-by-step explanation:
Solution: Given A.P. is 3, 15, 27, 39, …
first term, a = 3
common difference, d = a2 − a1 = 15 − 3 = 12
We know that,
an = a + (n − 1) d
Therefore,
a54 = a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636 = 639
a54 = 639
We have to find the term of this A.P. which is 132 more than a54, i.e. 771.
Let nth term be 771.
an = a + (n − 1) d
771 = 3 + (n − 1) 12
768 = (n − 1) 12
(n − 1) = 64
n = 65
Therefore, 65th term was 132 more than 54th term.
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