Question

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:

• A. 122
• B. 84
• C. 90
• D. 108

Hint:

To solve A.P. problems, use the formula for the sum of the first \(n\) terms and the properties of arithmetic progressions.

Answer 1

The Correct Option is D

Let the terms of the A.P. be \( a, a+d, a+2d, \dots \). - The sum of the first three terms is given by: \[ a + (a + d) + (a + 2d) = 54 \quad \Rightarrow \quad 3a + 3d = 54 \quad \Rightarrow \quad a + d = 18. \] - The sum of the first 20 terms is: \[ S_{20} = \frac{20}{2} \left( 2a + (20-1)d \right) = 10(2a + 19d). \] This lies between 1600 and 1800, so: \[ 1600 < 10(2a + 19d) < 1800 \quad \Rightarrow \quad 160 < 2a + 19d < 180. \] Substituting \( a + d = 18 \) into the equation gives \( a = 18 - d \), and solving for \( d \) gives the 11th term as \( 108 \). 
Thus, the 11th term is 108.