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. 108
• D. 90

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

We are tasked with solving the problem involving an arithmetic progression (AP) and determining the value of the 11th term \( a_{11} \). Let us proceed step by step:

1. Representation of the AP:
The terms of the AP are represented as:

\( a, a + d, a + 2d, \dots \)

2. Sum of the First Three Terms:
The sum of the first three terms is given as 54:

\( 3a + 3d = 54 \)

Dividing through by 3:

\( a + d = 18 \quad \text{(Equation i)} \)

3. Constraint on the Sum of the First 20 Terms:
The sum of the first 20 terms lies between 1600 and 1800:

\( 1600 < \frac{20}{2} [2a + 19d] < 1800 \)

Simplify the inequality:

\( 1600 < 10 [2a + 19d] < 1800 \)

Divide through by 10:

\( 160 < 2a + 19d < 180 \)

4. Substituting \( a + d = 18 \):
From Equation (i), \( a = 18 - d \). Substitute \( a = 18 - d \) into \( 2a + 19d \):

\( 2a + 19d = 2(18 - d) + 19d = 36 - 2d + 19d = 36 + 17d \)

Thus, the inequality becomes:

\( 160 < 36 + 17d < 180 \)

Subtract 36 from all sides:

\( 124 < 17d < 144 \)

Divide through by 17:

\( \frac{124}{17} < d < \frac{144}{17} \)

\( 7.29 < d < 8.47 \)

Since \( d \) must be an integer, we conclude:

\( d = 8 \)

5. Solving for \( a \):
From Equation (i), \( a + d = 18 \):

\( a + 8 = 18 \)

\( a = 10 \)

6. Finding \( a_{11} \):
The general formula for the \( n \)-th term of an AP is:

\( a_n = a + (n-1)d \)

For \( n = 11 \):

\( a_{11} = a + 10d \)

Substitute \( a = 10 \) and \( d = 8 \):

\( a_{11} = 10 + 10 \cdot 8 = 10 + 80 = 90 \)

Final Answer:
The value of \( a_{11} \) is \( \boxed{90} \).

Answer 2

Step 1: Given information.
Let the terms of the arithmetic progression (A.P.) be denoted as \( a, a+d, a+2d, \dots \), where \( a \) is the first term and \( d \) is the common difference.
We are given two conditions:
- The sum of the first three terms is 54.
- The sum of the first twenty terms lies between 1600 and 1800.
We need to find the 11th term of the A.P.

Step 2: Use the sum of the first three terms.
The sum of the first three terms is:
\[ S_3 = a + (a+d) + (a+2d) = 3a + 3d. \] We are given that this sum is 54:
\[ 3a + 3d = 54 \quad \Rightarrow \quad a + d = 18. \quad \text{(Equation 1)} \]

Step 3: Use the sum of the first twenty terms.
The sum of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right). \] For the first twenty terms, we have:
\[ S_{20} = \frac{20}{2} \left( 2a + 19d \right) = 10 \left( 2a + 19d \right). \] We are given that \( 1600 < S_{20} < 1800 \), so:
\[ 160 < 2a + 19d < 180. \quad \text{(Equation 2)} \]

Step 4: Solve the system of equations.
From Equation 1, \( a + d = 18 \), we can solve for \( a \):
\[ a = 18 - d. \] Substitute this into Equation 2:
\[ 160 < 2(18 - d) + 19d < 180. \] Simplifying:
\[ 160 < 36 - 2d + 19d < 180, \] \[ 160 < 36 + 17d < 180. \] Subtract 36 from all parts of the inequality:
\[ 124 < 17d < 144. \] Now, divide by 17:
\[ \frac{124}{17} < d < \frac{144}{17}, \] \[ 7.29 < d < 8.47. \] Since \( d \) is an integer, we conclude:
\[ d = 8. \]

Step 5: Find the value of \( a \).
Substitute \( d = 8 \) into Equation 1:
\[ a + 8 = 18 \quad \Rightarrow \quad a = 10. \]

Step 6: Find the 11th term of the A.P.
The 11th term of the A.P. is:
\[ T_{11} = a + 10d = 10 + 10 × 8 = 10 + 80 = 90. \]

Final Answer:
\[ \boxed{90}. \]