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  • The 4thterm from the end of the AP: –11, –8, –5, ..., 49 is (A) 37 (B) 40 (C) 43 (D) 58

The 4thterm from the end of the AP: –11, –8, –5, ..., 49 is

(A) 37

(B) 40

(C) 43

(D) 58

Answers (1)

\text{Given AP is } -11, -8, -5, \ldots, 49 \\ \text{Here, } a_{1} = -11, \quad a_{2} = -8, \quad a_{n} = 49 \\ d = a_{2} - a_{1} \\ = -8 - (-11) \\ = -8 + 11 \\ = 3 \\ \text{We know that } a_{n} = a_{1} + (n - 1) \cdot d \\ \text{Put } 49 = -11 + (n - 1)(3) \\ 49 = -11 + 3n - 3 \\ 49 = -14 + 3n \\ 49 + 14 = 3n \\ 63 = 3n \\ n = \frac{63}{3} \\ n = 21 \\ a_{n} = a_{1} + (n - 1) \cdot d \\ \text{Put } n = 21 - 3 = 18 \\ a_{18} = -11 + (18 - 1) \times 3 \\ a_{18} = -11 + 17 \times 3 \\ a_{18} = -11 + 51 \\ a_{18} = 40 \\ \text{40 is the 4th last term from end.}

Hence option B is correct

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