Question

The sum of the first six terms of an arithmetic progression is 54 and the ratio of the 10^th term to its 30^th term is 11 : 31. What is the 60^th term of the progression?

• A. 116
• B. 120
• C. 122
• D. 126

Hint:

For arithmetic progression problems, use the sum formula for the first \( n \) terms to establish relations between \( a \) and \( d \). Then, solve using the given ratios of terms or sums to find the unknowns.

Answer 1

The Correct Option is D

Let the first term of the arithmetic progression be \( a \) and the common difference be \( d \).
The sum of the first \( n \) terms is given by: \[ S_n = \frac{n}{2} \left( 2a + (n - 1) d \right) \] We are given that the sum of the first six terms is 54, so: \[ S_6 = \frac{6}{2} \left( 2a + 5d \right) = 54 \quad \Rightarrow \quad 3(2a + 5d) = 54 \quad \Rightarrow \quad 2a + 5d = 18 \quad \cdots (1) \] Also, the ratio of the 10\textsuperscript{th} term to the 30\textsuperscript{th} term is 11 : 31, so: \[ \frac{a + 9d}{a + 29d} = \frac{11}{31} \] Cross multiplying: \[ 31(a + 9d) = 11(a + 29d) \quad \Rightarrow \quad 31a + 279d = 11a + 319d \] \[ 20a = 40d \quad \Rightarrow \quad a = 2d \quad \cdots (2) \] Substituting \( a = 2d \) into equation (1): \[ 2(2d) + 5d = 18 \quad \Rightarrow \quad 4d + 5d = 18 \quad \Rightarrow \quad 9d = 18 \quad \Rightarrow \quad d = 2 \] Thus, \( a = 2 \times 2 = 4 \). Now, the 60\textsuperscript{th} term is given by: \[ T_{60} = a + 59d = 4 + 59 \times 2 = 4 + 118 = 122 \] Thus, the 60\textsuperscript{th} term is 122.