Question

For any natural number n,suppose the sum of the first n terms of an arithmetic progression is \((n+2n^2)\). If the \(n^{th}\) term of the progression is divisible by 9,then the smallest possible value of n is

• A. 4
• B. 8
• C. 7
• D. 9

Answer 1

The Correct Option is C

The correct answer is C:7
Given information:
Sum of the first n terms of an arithmetic progression \((AP) = n + 2n^2\)
nth term of the AP is divisible by 9.
Step 1: Express the Sum of n Terms
The sum of the first n terms of an arithmetic progression can be expressed using the formula: \(S_n = \frac{n}{2}\times[2a + (n - 1)d]\), where a is the first term and d is the common difference.
In this case, we have: \(n + 2n^2 = \frac{n}{2}\times[2a + (n - 1)d]\)
Step 2: Simplify the Equation
Simplify the equation: \(n + 2n^2 = \frac{n}{2}\times[2a + (n - 1)d]\)
Divide both sides by n to get: \(1+2n=\frac{1}{2}\times[2a+(n-1)d]\)
Step 3: Analyze Divisibility by 9
For the nth term of the AP to be divisible by 9, the expression 2a+(n-1)d must be divisible by 9.
Step 4: Consider Cases for 2a+(n-1)d
We consider different cases for 2a+(n-1)d to be divisible by 9:
Case 1: 2a+(n-1)d = 9
In this case, n can be any natural number greater than 1.
Case 2: 2a+(n-1)d=18
In this case, we have 2a+(n-1)d=18.Since 2a and (n-1)d are both positive integers, the smallest possible value for n is 7 (when 2a=2 and (n-1)d= 16).
Step 5: Determine the Smallest Possible Value of n
Among the cases considered, the smallest possible value of n is 7 (from Case 2).
Therefore, the smallest possible value of n is indeed 7.

Answer 2

Given, \(S_n = (n + 2n^2)\)
Let \(𝑇_𝑛\) stand for the \(𝑛^{𝑡ℎ}\) term of the same arithmetic progression.
\(T_n = S_n - S_{n-1}\)
\(T_n = (n + 2n^2) - (n-1 + 2(n-1)^2)\)
\(T_n = (n + 2n^2) - (n-1 + 2(n^2+1-2n))\)
\(T_n = (n + 2n^2) - (n-1 + 2n^2+2-4n)\)
\(T_n = (n + 2n^2 - n + 1 - 2n^2 - 2 + 4n)\)
\(T_n = 4n - 1\)
The series' first term that is divisible by nine is 27. 
The seventh period is 27. 
Consequently, 7 is the lowest value of n that can exist. 
Option C is the solution.