Which term of the A.P, 3, 15, 27, 39, ... will be 132 more than its 54th term.
Answers
Answered by
1
Answer:A=3
D=15-3=12
an=a+(n-1).d
132=3+(n-1).12
132=3+12n-12
132-3=12n-12
129=12n-12
129+12=12n
141=12n
n=11.5
Step-by-step explanation:
Answered by
1
Solution
65th term will be 132 more than the 54th term
Step-by-step explanation:
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
Similar questions