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:

Question

Let $a_1,a_2,a_3,a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1^2a_2a_3a_4+1^4=361$, then the largest term of the A.P. is equal to

Answer 1

The objective is to determine the 11th term ($ a_{11} $) of an arithmetic progression (AP). The solution proceeds as follows:

1. AP Representation:
An AP is defined by its first term ($ a $) and common difference ($ d $), with terms given by $ a, a + d, a + 2d, \dots $.

2. Sum of First Three Terms:
Given that the sum of the first three terms is 54:

$ a + (a + d) + (a + 2d) = 54 $

$ 3a + 3d = 54 $

Dividing by 3 yields:

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

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

$ 1600 < S_{20} < 1800 $

Using the formula $ S_n = \frac{n}{2} [2a + (n-1)d] $:

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

$ 1600 < 10 [2a + 19d] < 1800 $

Dividing by 10:

$ 160 < 2a + 19d < 180 $

4. Substituting Equation (i):
From Equation (i), $ a = 18 - d $. Substituting this into the inequality:

$ 2(18 - d) + 19d = 36 - 2d + 19d = 36 + 17d $

The inequality becomes:

$ 160 < 36 + 17d < 180 $

Subtracting 36 from all parts:

$ 124 < 17d < 144 $

Dividing by 17:

$ \frac{124}{17} < d < \frac{144}{17} $

$ 7.29 < d < 8.47 $

Since $ d $ must be an integer, $ d = 8 $.

5. Solving for $ a $:
Using Equation (i), $ a + d = 18 $:

$ a + 8 = 18 $

$ a = 10 $

6. Calculating $ a_{11} $:
The formula for the $ n $-th term is $ a_n = a + (n-1)d $. For the 11th term ($ n = 11 $):

$ a_{11} = a + (11-1)d = a + 10d $

Substituting $ a = 10 $ and $ d = 8 $:

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

Final Result:
The 11th term of the AP is $ \boxed{90} $.

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:

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The Correct Option is D

Solution and Explanation

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