Question

Find the 10th term from the end of the AP 4, 9, 14, ..., 254.
Or, Which term of the AP 3, 15, 27, 34, ... will be 132 more than its 54th term?

Answer 1

Here, a = 4, d = (9 − 4) = 5, l = 254 and n = 10
Now, n^th term from the end = {l − (n − 1)d}
∴ 10^th term from the end = {254 − (10 − 1) × 5}
= {254 − (9 × 5)} = (254 − 45) = 209
Hence, the 10^th term from the end is 209.

OR
Here, a = 3, d = (15 − 3) = 12
T_n = 132 + T_⁵⁴
⇒ a + (n − 1)d = 132 + {3 + (54 − 1) × 12}
⇒ 3 + (n − 1) × 12 = 132 + 639
⇒(n − 1) × 12 = (771 − 3) = 768
⇒ (n − 1) = $\frac{768}{12}=64$
∴ n = (64 + 1) = 65
Hence, the required term is the 65^th term.