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Question
Which term of the AP 3, 15, 27, 39, ...... will be 120 more than its 21st term?
Solution
Given AP is
3, 15, 27, 39 .....
where a = 3, d = 15 - 3 = 12
Let the nth term be 120 more than its 21st term.
tn = t21 + 120
⇒ 3 +(n -1 )12 = 3 +20 × 12 + 120
⇒ (n - 1) × 12 = 363 - 3
⇒ (n - 1) =`360/12`
∴ n = 31
Hence , the required term is
t31 = 3 + 30 × 12
= 363
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