Question

which term of the AP 3 15 27 39 will be 132 more than its 54th term

Answer 1

GIVEN
54th term
a + 53 d
132 more than a54
a + 53d + 132
Therefore, this is your an (last term)
AP 3 15 27 39,...
first term 'a' = 3
Common difference 'd' = 15 - 3 = 12
an = a + 53d + 132
FORMULA
an = a + (n - 1)d
a + 53d + 132 = a + (n - 1)d
Subsitute the values of a and d
3 + 53x12 + 132 = 3 + ( n - 1) x 12
3 + 53x12 + 132 - 3 = 12( n - 1)
636 + 132 = 12( n - 1)
768 = 12( n - 1)
768 / 12 = n - 1
64 = n - 1
64 + 1 = n
n = 65
Therefore, the required term is 65th.

Answer 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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