Question

The sum of first n terms of an A.P. is given by \(S_n = 4n^2 - n\). Find the 25th term of this A.P.

• A. 195
• B. 198
• C. 203
• D. 190

Hint:

If \(S_n = An^2 + Bn\), then the \(n\)-th term of the A.P. is given by the general shortcut formula:
\[ a_n = 2An + (B - A) \]
For \(S_n = 4n^2 - n\), we have \(A = 4\) and \(B = -1\).
Substituting these values:
\[ a_n = 2(4)n + (-1 - 4) = 8n - 5 \]
Now, simply substitute \(n = 25\) to get the term:
\[ a_{25} = 8(25) - 5 = 200 - 5 = 195 \]
This shortcut is extremely powerful for quick MCQs!

Answer 1

The Correct Option is A

Step 1: Understand what is being asked.
The sum of the first $n$ terms is $S_n=4n^2-n$, and we need the 25th term. Instead of computing two specific sums, let us derive a general formula for the $n$-th term in terms of $n$, and then substitute $n=25$.

Step 2: Write the general expression for the $n$-th term.
For any A.P, the $n$-th term equals $S_n-S_{n-1}$. Let us compute this symbolically:
\[ S_{n-1} = 4(n-1)^2-(n-1) \]
Expand $(n-1)^2=n^2-2n+1$:
\[ S_{n-1} = 4(n^2-2n+1)-(n-1) = 4n^2-8n+4-n+1 = 4n^2-9n+5 \]
Step 3: Subtract to get $a_n$ as a general formula.
\[ a_n = S_n - S_{n-1} = (4n^2-n)-(4n^2-9n+5) \]
\[ a_n = 4n^2-n-4n^2+9n-5 = 8n-5 \]
Step 4: Substitute $n=25$ into this general formula.
\[ a_{25} = 8(25)-5 = 200-5 = 195 \]
Final Answer:
The 25th term of the A.P. is 195, matching option (A).
\[ \boxed{195} \]