Or, which term of the AP 3, 15, 27, 34, … will be 132 more than its 54th term?
Which term of the AP 3, 15, 27, 39, … will be 132 more than its 54th term.
Given AP = 3, 15, 27, 39, …
First term, a = 3
Common difference, a2 - a1 = 15 - 3 = 12
And we know
Nth term of an AP, an = a + (n - 1)d
Where a is first term and d is common difference.
Now, let the mth term be 132 more than 54th term
In that case,
am = a54 + 132
⇒ a + (m - 1)d = a + 53d + 132
⇒ (m - 1)12 = 53(12) + 132
⇒ 12m - 12 = 636 + 132
⇒ 12m = 768 + 12
⇒ 12m = 780
⇒ m = 65
hence, 65th term will be 132 more than its 54th term.