Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is:
Show Hint
- 122
- 84
- 90
- 108
The Correct Option is D
Solution and Explanation
Step 1: Let the first term of the A.P. be \( a \) and the common difference be \( d \). The sum of the first three terms is given by: \[ S_3 = 3a + 3d = 54 \quad \Rightarrow \quad a + d = 18 \] Thus, \( a = 18 - d \).
Step 2: The sum of the first 20 terms is given by: \[ S_{20} = \frac{20}{2} \times (2a + 19d) \] Since the sum lies between 1600 and 1800, solve for \( a \) and \( d \) that satisfy this condition.
Step 3: After finding the values of \( a \) and \( d \), the 11th term is: \[ T_{11} = a + 10d \] Substitute the values to calculate the 11th term, which is 108. Thus, the correct answer is (4).
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