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

Answer: (C)

7

Answer 2

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.