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

To solve this problem, we start by recalling the formula for the sum of the first \( n \) terms of an arithmetic progression (AP), which is given by:

\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

Given that the sum of the first \( n \) terms is \((n + 2n^2)\), we equate this with the formula:

\[ \frac{n}{2} \left(2a + (n-1)d\right) = n + 2n^2 \]

Simplifying, we have:

\[ n(2a + (n-1)d) = 2n + 4n^2 \]

Dividing through by \( n \) (since \( n \neq 0 \)), we get:

\[ 2a + (n-1)d = 2 + 4n \]

Now, the nth term of the AP, denoted as \( T_n \), is defined by:

\[ T_n = a + (n-1)d \]

To find the nth term, we rearrange the equation:

\[ 2a + (n-1)d = 2 + 4n \]

Let’s solve for \( a + (n-1)d \):

Substitute \( 2a \) in terms of \((2a + (n-1)d) \):

\[ a + (n-1)d = \frac{2 + 4n + 2}{2} \]

\[ a + (n-1)d = 2 + 2n \]

Thus the nth term is \( 2 + 2n \).

According to the question, this nth term is divisible by 9:

\[ 2 + 2n \equiv 0 \pmod{9} \]

Rearranging gives:

\[ 2n \equiv -2 \equiv 7 \pmod{9} \]

To solve for \( n \), multiply both sides by the modular inverse of 2 mod 9, which is 5 (since \( 2 \times 5 \equiv 1 \pmod{9} \)):

\[ 10n \equiv 35 \pmod{9} \]

\[ n \equiv 35 \pmod{9} \]

Calculating \( 35 \mod 9 \) gives:

\[ 35 \div 9 = 3 \text{ R } 8 \]

\[ n \equiv 8 \pmod{9} \]

Thus, the smallest possible value of \( n \) is:

7

Answer 2

Given: \(S_n = n + 2n^2\) 

We are asked to find the first term of the series divisible by 9. Let \(T_n\) be the \(n^\text{th}\) term of the sequence.

The \(n^\text{th}\) term of a sequence defined by its sum formula is: \(T_n = S_n - S_{n-1}\)

So, \(T_n = (n + 2n^2) - \left[(n - 1) + 2(n - 1)^2\right]\)

Now simplify the expression:
First, expand \((n - 1)^2 = n^2 - 2n + 1\)
So, \(T_n = (n + 2n^2) - \left[(n - 1) + 2(n^2 - 2n + 1)\right]\)

Expand the second bracket: \((n - 1) + 2(n^2 - 2n + 1) = n - 1 + 2n^2 - 4n + 2 = 2n^2 - 3n + 1\)

Now subtract: \(T_n = n + 2n^2 - (2n^2 - 3n + 1)\)

Simplify: \(T_n = n + 2n^2 - 2n^2 + 3n - 1 = 4n - 1\)

So the general term of the sequence is: \(T_n = 4n - 1\)

Now, to find the first term divisible by 9, set: \(4n - 1 \equiv 0 \mod 9\)
\(4n \equiv 1 \mod 9\)

Multiply both sides by the modular inverse of 4 mod 9.
Since \(4 \times 7 = 28 \equiv 1 \mod 9\), the inverse is 7.

So, \(n \equiv 7 \mod 9\)

Therefore, the smallest such \(n\) is: \(\boxed{7}\)

Check: \(T_7 = 4 \times 7 - 1 = 28 - 1 = 27\), and 27 is divisible by 9.

Correct option: C