nth Term from End of AP
In many problems, the nth term from the end of an arithmetic progression (AP) is required instead of the nth term from the beginning. This occurs when the last term is given but the first term is unknown, or when counting backwards is more efficient.
The nth term from the end can be found using a simple formula that treats the AP as if it were reversed — starting from the last term and subtracting the common difference repeatedly.
This is a frequently tested concept in CBSE Class 10 board exams and appears in both MCQ and short-answer questions.
What is nth Term from End of AP?
Definition: The nth term from the end of a finite AP is the term obtained by counting n positions backward from the last term.
If the AP is: a, a+d, a+2d, ..., l (where l is the last term)
- 1st term from end = l
- 2nd term from end = l − d
- 3rd term from end = l − 2d
- nth term from end = l − (n−1)d
Key insight: An AP read backwards is also an AP with:
- First term = l (the last term of the original AP)
- Common difference = −d (negative of the original common difference)
nth Term from End of AP Formula
nth Term from the End:
nth term from end = l − (n − 1)d
Where:
- l = last term of the AP
- n = position from the end (1st, 2nd, 3rd, ...)
- d = common difference of the AP
Alternative (using total terms):
If the AP has N terms total, then the nth term from the end is the (N − n + 1)th term from the beginning.
nth from end = a + (N − n)d
Where:
- a = first term
- N = total number of terms
Derivation and Proof
Derivation of the formula:
- Consider the AP: a, a+d, a+2d, ..., l.
- The reversed AP is: l, l−d, l−2d, ..., a.
- This reversed AP has first term = l and common difference = −d.
- The nth term of the reversed AP = l + (n−1)(−d) = l − (n−1)d.
- This is exactly the nth term from the end of the original AP.
Equivalence with the other formula:
- The nth term from the end is the (N−n+1)th term from the beginning.
- Using aₖ = a + (k−1)d with k = N−n+1:
- = a + (N−n+1−1)d = a + (N−n)d
- Since l = a + (N−1)d, we get a = l − (N−1)d.
- Substituting: = [l − (N−1)d] + (N−n)d = l − (n−1)d. Verified.
Types and Properties
Types of problems involving nth term from end:
- Direct computation: Given a, d, l (or N), find the nth term from the end.
- Finding which term from end: Given a specific value, determine its position from the end.
- Comparing terms from beginning and end: Problems like "the 4th term from end equals the 10th term from beginning."
- Finding the middle term: In an AP with N terms, the middle term is the ((N+1)/2)th term from both the beginning and the end (when N is odd).
Methods
Method 1: Using l − (n−1)d (when last term is given):
- Identify the last term l and common difference d.
- Apply: nth term from end = l − (n−1)d.
Method 2: Using a + (N−n)d (when total terms and first term are given):
- Identify first term a, common difference d, total terms N.
- Find the position from the beginning: k = N − n + 1.
- Apply: aₖ = a + (k−1)d = a + (N−n)d.
Method 3: Find total terms first, then convert:
- If l, a, d are given but N is not, find N using: l = a + (N−1)d → N = (l−a)/d + 1.
- Then position from beginning = N − n + 1.
- Find that term.
Common mistake:
- Using l − nd instead of l − (n−1)d. The 1st term from end is l itself (not l − d), so we subtract (n−1)d, not nd.
Solved Examples
Example 1: Direct Computation
Problem: Find the 5th term from the end of the AP: 3, 7, 11, ..., 103.
Solution:
Given:
- l = 103, d = 7 − 3 = 4, n = 5
Using: nth term from end = l − (n−1)d
- = 103 − (5−1)(4)
- = 103 − 16
- = 87
Answer: 5th term from end = 87
Example 2: Finding 3rd Term from End
Problem: Find the 3rd term from the end of the AP: 5, 9, 13, ..., 89.
Solution:
Given:
- l = 89, d = 9 − 5 = 4, n = 3
Using the formula:
- = 89 − (3−1)(4)
- = 89 − 8
- = 81
Answer: 3rd term from end = 81
Example 3: Using Total Terms Method
Problem: The AP has first term 2, common difference 5, and 20 terms. Find the 4th term from the end.
Solution:
Given:
- a = 2, d = 5, N = 20, n = 4
Method 1 (using position):
- Position from beginning = N − n + 1 = 20 − 4 + 1 = 17
- a₁₇ = 2 + (17−1)(5) = 2 + 80 = 82
Method 2 (using last term):
- l = a + (N−1)d = 2 + 19(5) = 97
- 4th from end = 97 − (4−1)(5) = 97 − 15 = 82
Answer: 4th term from end = 82
Example 4: Term from End in Decreasing AP
Problem: Find the 6th term from the end of the AP: 50, 47, 44, ..., 5.
Solution:
Given:
- l = 5, d = 47 − 50 = −3, n = 6
Using the formula:
- = 5 − (6−1)(−3)
- = 5 − (−15)
- = 5 + 15
- = 20
Verification: AP is 50, 47, 44, ..., 20, 17, 14, 11, 8, 5. The 6th from end is indeed 20.
Answer: 6th term from end = 20
Example 5: Comparing Terms from Both Ends
Problem: In an AP, the 4th term from the end is 20 and the 4th term from the beginning is 8. If d = 2, find the total number of terms.
Solution:
Given:
- a₄ = 8 → a + 3d = 8 → a + 6 = 8 → a = 2
- 4th from end = 20 → l − 3d = 20 → l − 6 = 20 → l = 26
Finding N:
- l = a + (N−1)d
- 26 = 2 + (N−1)(2)
- 24 = 2(N−1)
- N − 1 = 12
- N = 13
Answer: Total number of terms = 13
Example 6: Finding Position from End
Problem: In the AP: 7, 13, 19, ..., 205, which term from the end is 157?
Solution:
Given:
- l = 205, d = 13 − 7 = 6
- nth term from end = l − (n−1)d
Setting up equation:
- 157 = 205 − (n−1)(6)
- (n−1)(6) = 205 − 157 = 48
- n − 1 = 8
- n = 9
Answer: 157 is the 9th term from the end.
Example 7: Middle Term of an AP
Problem: Find the middle term of the AP: 10, 15, 20, ..., 100.
Solution:
Finding N:
- a = 10, d = 5, l = 100
- 100 = 10 + (N−1)(5)
- 90 = 5(N−1)
- N − 1 = 18, N = 19
Middle term position: (19+1)/2 = 10th term
- a₁₀ = 10 + 9(5) = 10 + 45 = 55
Verification from end: 10th from end = 100 − 9(5) = 100 − 45 = 55 ✔
Answer: Middle term = 55
Example 8: Finding AP Given Terms from Both Ends
Problem: In a finite AP, the 7th term from the beginning equals the 7th term from the end. If the AP has 13 terms, prove that the 7th term is the average of the first and last terms.
Solution:
Given: N = 13
- 7th from beginning = a + 6d
- 7th from end = l − 6d
Setting them equal:
- a + 6d = l − 6d
- 12d = l − a
- Since l = a + 12d (for 13 terms), this gives l − a = 12d ✔
The 7th term:
- a₇ = a + 6d
- Since l = a + 12d, we get d = (l−a)/12
- a₇ = a + 6 × (l−a)/12 = a + (l−a)/2 = (2a + l − a)/2 = (a + l)/2
Answer: The 7th term = (a + l)/2, which is the average of the first and last terms. This is because in an AP with an odd number of terms, the middle term always equals (a + l)/2.
Real-World Applications
Counting Backward in Sequences:
- Finding recent terms in time-based sequences (monthly savings, depreciation schedules).
Finding Middle Terms:
- The middle term of an AP with odd number of terms equals (a + l)/2. This uses the concept of counting from both ends.
Competition Mathematics:
- Many olympiad problems require relating terms from the beginning and end of arithmetic sequences.
Data Analysis:
- In evenly spaced data (like hourly temperatures), the nth reading from the end can be quickly calculated without listing all values.
Key Points to Remember
- nth term from end = l − (n−1)d, where l is the last term.
- Alternatively: nth from end = a + (N−n)d, where N is total terms.
- The 1st term from end is l itself (n=1 gives l − 0 = l).
- For decreasing APs (d < 0), subtracting a negative d means adding.
- An AP read backwards is also an AP with first term l and common difference −d.
- The nth term from end = (N−n+1)th term from beginning.
- Middle term of an AP with N terms (N odd) = (a + l)/2.
- In an AP with N terms, the kth term from beginning + the kth term from end = a + l (constant sum).
- Common mistake: using l − nd instead of l − (n−1)d.
- This is a short but high-scoring topic in CBSE Class 10 exams.
Practice Problems
- Find the 4th term from the end of the AP: 2, 6, 10, ..., 98.
- Find the 7th term from the end of the AP: 1, 4, 7, ..., 88.
- The AP has first term 5, last term 95, and common difference 3. Find the 10th term from the end.
- In the AP: 100, 95, 90, ..., 10, find the 8th term from the end.
- Find the middle term of the AP: 3, 8, 13, ..., 253.
- In an AP, the 5th term from end is 32 and the 5th term from beginning is 12. If d = 5, find the number of terms.
- Which term from the end is 27 in the AP: 3, 5, 7, ..., 51?
Frequently Asked Questions
Q1. What is the formula for the nth term from the end of an AP?
The nth term from the end = l − (n−1)d, where l is the last term and d is the common difference. For example, the 3rd term from the end of the AP 5, 10, 15, ..., 100 is 100 − 2(5) = 90.
Q2. How is the nth term from end related to the nth term from beginning?
In an AP with N terms, the nth term from the end is the (N − n + 1)th term from the beginning. So the 3rd from end in a 20-term AP is the 18th from beginning.
Q3. Why do we subtract (n-1)d and not nd?
Because the 1st term from the end is the last term itself (l), not l − d. We count back (n−1) steps to reach the nth term from end. If we subtracted nd, we would overshoot by one position.
Q4. What if the AP has a negative common difference?
The formula still works. If d = −3, the nth term from end = l − (n−1)(−3) = l + 3(n−1). In a decreasing AP, terms from the end are larger than the last term (moving backward means going up).
Q5. How do I find the middle term?
For an AP with N terms (N must be odd for a single middle term): middle term is the ((N+1)/2)th term from either end. Its value equals (a + l)/2, the average of the first and last terms.
Q6. What if both the first and last terms are given but d is not?
If you know a, l, and N, find d = (l − a)/(N − 1). Then apply the nth-from-end formula. If N is also unknown, you need one more piece of information.
Q7. Is this formula in the NCERT textbook?
The formula l − (n−1)d is derived from the general term formula and the concept of reversing an AP. NCERT discusses it as a direct application of the nth term formula. It regularly appears in CBSE board exam questions.
Q8. What is the sum of the kth term from the beginning and the kth term from the end?
In any finite AP, this sum is constant and equals a + l (first term + last term). This is because: [a + (k−1)d] + [l − (k−1)d] = a + l. This property is used in the sum formula.
