which term of AP 3,15,27,39... will be 132 more than its 54th term?
Answers
Answered by
24
Hi,
Here is your answer,
Given data,
a(First term) = 3 , d(common difference) = 12 , n = 54
Apply the formula
⊕ a+(n -1) × d
→ 3 + (54 - 1) × 12
→ 3+53×12
→ 3+636
→ 639
Final answer + 132
→ 132 + 639 = 771
→ 771 → 3+(n - 1) × 12
→ 768/12 + 1 → n
→ n = 64
→ n = 64 + 1
→ n = 65
Hope it helps you !
Here is your answer,
Given data,
a(First term) = 3 , d(common difference) = 12 , n = 54
Apply the formula
⊕ a+(n -1) × d
→ 3 + (54 - 1) × 12
→ 3+53×12
→ 3+636
→ 639
Final answer + 132
→ 132 + 639 = 771
→ 771 → 3+(n - 1) × 12
→ 768/12 + 1 → n
→ n = 64
→ n = 64 + 1
→ n = 65
Hope it helps you !
chandanshetty:
thank you bro
Answered by
2
☺ Hello mate__ ❤
◾◾here is your answer...
Lets first calculate 54th of the given AP.
First term = a = 3
Common difference = d =15 - 3 = 12
Using formula an=a+(n−1)d, to find nth term of arithmetic progression, we get
a54=a+(54−1)d
a54=3+53(12)=3+636=639
We want to find which term is 132 more than its 54th term. Lets suppose it is nth term which is 132 more than 54th term.
Therefore, we can say that
an=a54+132
......{an=a+(n−1)d=3+(n−1)(12)}{a54=639}
⇒3+(n−1)12=639+132
⇒3+12n−12=771
⇒12n−9=771
⇒12n=780
⇒n=780/12=65
Therefore, 65th term is 132 more than its 54th term.
I hope, this will help you.
Thank you______❤
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