Question

The mean of \(n\) terms of an arithmetic progression is \(\bar{x}\). If the sum of \((n-1)\) terms is \(k\), find the \(n\)th term.

• A. \(n\bar{x}+k\)
• B. \(\bar{x}-nk\)
• C. \(k\bar{x}+n\)
• D. \(n\bar{x}-k\)

Hint:

Sum of n terms is \(n\bar{x}\); the nth term equals that sum minus the sum of the first \((n-1)\) terms.

Answer 1

The Correct Option is D

Step 1: Connect the mean to the total sum.
The mean of any set of numbers is the sum divided by how many numbers there are. So if \(\bar{x}\) is the mean of \(n\) terms, the sum of all \(n\) terms is
\[ S_n = n\bar{x} \]

Step 2: Recall how a single term connects to two sums.
For any sequence, the last (nth) term equals the sum of the first \(n\) terms minus the sum of the first \((n-1)\) terms, because removing the first \((n-1)\) terms from the full sum leaves only the nth term behind.
\[ T_n = S_n - S_{n-1} \]

Step 3: Substitute the known values.
We are told the sum of \((n-1)\) terms is \(k\), so \(S_{n-1}=k\). Using \(S_n=n\bar{x}\) from Step 1:
\[ T_n = n\bar{x} - k \]

Step 4: Verify with a quick example.
Take the AP \(2, 4, 6, 8, 10\), so \(n=5\). The mean is \(\bar{x}=\frac{2+4+6+8+10}{5}=\frac{30}{5}=6\). The sum of the first \((n-1)=4\) terms is \(k=2+4+6+8=20\). The actual 5th term is \(10\). Checking the formula: \(n\bar{x}-k = 5(6)-20 = 30-20=10\), which matches the real 5th term exactly.

Step 5: Why the other options are wrong.
Option (a), \(n\bar{x}+k\), adds instead of subtracts, which would double-count part of the sum. Option (b), \(\bar{x}-nk\), mixes up which quantity gets multiplied by \(n\). Option (c), \(k\bar{x}+n\), multiplies two unrelated quantities together and does not come from any valid sum relation. Testing any of these against the worked example above gives the wrong 5th term.

Final Answer:
The nth term is \(n\bar{x}-k\), option (d). \[ \boxed{n\bar{x}-k} \]