which term of A.P 3,15,27,39... will be 132 more than its 54th term?
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We know that an=a+(n-1) * d ---------- (1)
= 3 + (54-1) * 12
= 3 + 53 * 12
= 639.
a54 = 639.
Given that the term is 132 more than its 54th term = 639 + 132 = 771.
on substituting in (1), we get
771 = 3 + (n-1) * 12
768 = (n-1) * 12
768/12 = n-1
n-1 = 64
n = 65.
Therefore the 65th term is 132 more than its 54th term.
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