Exercise 6.2 — nth Term of AP

General (nth) term of an arithmetic progression.

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Exercise 6.2 — The nth Term of an Arithmetic Progression

Exercise 6.2 of Chapter 6, Progressions (Class 10 Mathematics — CBSE, Telangana & Andhra Pradesh syllabus), introduces the single most useful tool in the whole chapter: the nth-term formula. Instead of writing out an AP term by term, this formula lets you jump straight to any term — the 30th, the 100th, even the 1000th — in one calculation.

This is the longest exercise in the chapter, with 17 questions ranging from direct table-filling and term-finding to word problems about salaries, three-digit numbers divisible by 7, and comparing two different APs. Every single one of them is solved using the same core formula, applied in different directions.

nth Term Formula Finding Missing Values Locating a Term Word Problems Two-AP Comparisons

💡 The one formula that solves this entire exercise: Every question in Exercise 6.2 is just this equation rearranged for a different unknown — sometimes solving for aₙ, sometimes for a, sometimes for d, and sometimes for n itself.

Deriving the nth Term Formula

Recall the general form of an AP: a, a+d, a+2d, a+3d, a+4d, ... Writing out the first few terms with their position numbers reveals a clear pattern in how many "d's" get added each time:

The position number is always one more than the number of times d has been added — because the first term needs zero additions. This gives the general rule:

aₙ = a + (n − 1)d

a = the first term of the AP
d = the common difference
n = the position of the term you want (1st, 2nd, 30th, 100th...)
aₙ = the value of that nth term
If an AP has a fixed number of terms, the very last term is also written as l (for "last term"), and l = a + (n−1)d where n is the total count of terms.

📌 One formula, four unknowns: The equation aₙ = a + (n−1)d has four quantities in it: a, d, n, and aₙ. Give any three, and you can always solve for the fourth — this single idea is the key to every question in this exercise.

Question 1 — Fill in the Blanks Using aₙ = a + (n−1)d

Each row of this table gives you three of the four quantities (a, d, n, aₙ) and asks you to find the missing one. The orange cells below show what was missing in each row, and the value that was found:

S.No.	a	d	n	aₙ
(i)	7	3	8	28
(ii)	−18	2	10	0
(iii)	46	−3	18	−5
(iv)	−18.9	2.5	10	3.6
(v)	3.5	0	105	3.5

Worked Example — Part (i)

a = 7, d = 3, n = 8 → find a₈

aₙ = a + (n−1)d a₈ = 7 + (8−1)(3) a₈ = 7 + 7(3) = 7 + 21 a₈ = 28

Worked Example — Part (iv)

a = −18.9, d = 2.5, aₙ = 3.6 → find n

3.6 = −18.9 + (n−1)(2.5) 3.6 + 18.9 = (n−1)(2.5) 22.5 = (n−1)(2.5) n − 1 = 22.5 ÷ 2.5 = 9 n = 9 + 1 = 10

📌 On part (v): Whenever d = 0, every term of the AP is identical to the first term — so a₁₀₅ is simply 3.5 again, no matter how large n is. There's no calculation needed once you spot d = 0.

Question 2 — Finding a Specific Term of an AP

Here, the AP is given as a list rather than as a, d values directly — so the first step in both parts is to read off a and calculate d = a₂ − a₁ before applying the nth-term formula.

Part (i)

Find the 30th term of the AP: 10, 7, 4, ...

a = 10, d = a₂ − a₁ = 7 − 10 = −3 a₃₀ = a + (30−1)d = 10 + 29(−3) a₃₀ = 10 − 87 = −77

Part (ii)

Find the 11th term of the AP: −3, −1/2, 2, ...

a = −3, d = a₂ − a₁ = −1/2 − (−3) = −1/2 + 3 = 5/2 a₁₁ = a + (11−1)d = −3 + 10 × 5/2 a₁₁ = −3 + 25 = 22

Question 3 — Finding the Missing Terms Between Two Known Terms

In this question, you're given two terms of an AP (not necessarily consecutive) and asked to find the terms in between or around them. The method has two flavours: when the first term is already known, one equation is enough; when neither given term is the first term, you need to solve two equations simultaneously.

Part	Given	d	Terms Found
(i)	a₁ = 2, a₃ = 26	12	a₂ = 14
(ii)	a₂ = 13, a₄ = 3	−5	a₁ = 18, a₃ = 8
(iii)	a₁ = 5, a₄ = 9½	3/2	a₂ = 6½, a₃ = 8
(iv)	a₁ = −4, a₆ = 6	2	a₂=−2, a₃=0, a₄=2, a₅=4
(v)	a₂ = 38, a₆ = −22	−15	a₁=53, a₃=23, a₄=8, a₅=−7

Worked Example — Part (i): When a₁ is already known

a₁ = 2, a₃ = 26 → find a₂

a₃ = a + 2d → 26 = 2 + 2d 2d = 26 − 2 = 24 → d = 12 a₂ = a + d = 2 + 12 = 14

Worked Example — Part (v): When neither term is a₁ (simultaneous equations)

a₂ = 38, a₆ = −22 → find a₁, a₃, a₄, a₅

Since a₁ isn't given directly, express it in terms of a₂ first, then substitute into the equation for a₆:

a₂ = a + d = 38 → a = 38 − d (equation 1) a₆ = a + 5d = −22 Substituting a = 38 − d: (38 − d) + 5d = −22 38 + 4d = −22 → 4d = −60 → d = −15 a = 38 − (−15) = 53 a₃ = a+2d = 53−30 = 23 a₄ = a+3d = 53−45 = 8 a₅ = a+4d = 53−60 = −7

⚠️ Don't assume the smaller subscript is a₁: In parts (ii) and (v), a₁ is not given — it has to be calculated. A common mistake is treating the first given term as though it were the first term of the AP.

Questions 4–9 — Locating a Term and Finding the Common Difference

This group of six questions all use the nth-term formula to answer a "which one / how many / is it even there" type question. The technique is always the same: set up aₙ = (the value asked about), then solve for whichever letter is unknown.

Question 4

Which term of the AP 3, 8, 13, 18, ... is 78?

a = 3, d = 5. Let the nth term = 78 3 + (n−1)(5) = 78 → (n−1)(5) = 75 → n−1 = 15 n = 16

✅ 78 is the 16th term of this AP.

Question 5

Find the number of terms in each AP

(i) 7, 13, 19, ..., 205 — a=7, d=6 7 + (n−1)(6) = 205 → n−1 = 198/6 = 33 → n = 34 (ii) 18, 15½, 13, ..., −47 — a=18, d=−5/2 18 + (n−1)(−5/2) = −47 → (n−1)(−5/2) = −65 → n−1 = 26 → n = 27

Question 6

Check whether −150 is a term of the AP: 11, 8, 5, 2, ...

a = 11, d = −3. Suppose −150 is the nth term: 11 + (n−1)(−3) = −150 → (n−1)(−3) = −161 n − 1 = 161/3 = 53.67 (not a whole number)

⛔ Conclusion: Since n must be a positive whole number (a term position can't be "the 53.67th term"), −150 is not a term of this AP.

Question 7

Find the 31st term of an AP whose 11th term is 38 and 16th term is 73

a + 10d = 38 (eq. 1) a + 15d = 73 (eq. 2) Subtracting (1) from (2): 5d = 35 → d = 7 From (1): a + 10(7) = 38 → a = 38 − 70 = −32 a₃₁ = a + 30d = −32 + 30(7) = −32 + 210 = 178

Question 8

If the 3rd term is 4 and the 9th term is −8, which term is zero?

a + 2d = 4 (eq. 1) a + 8d = −8 (eq. 2) Subtracting: 6d = −12 → d = −2 From (1): a + 2(−2) = 4 → a = 8 Set aₙ = 0: 8 + (n−1)(−2) = 0 → (n−1) = 4 → n = 5

✅ The 5th term of this AP is zero.

Question 9

The 17th term exceeds the 10th term by 7. Find the common difference.

a₁₇ = a₁₀ + 7 (a + 16d) = (a + 9d) + 7 16d − 9d = 7 → 7d = 7 d = 1

Question 10 — Two APs with the Same Common Difference

Two APs share the same common difference d, but have different first terms — call them x and y. We're told their 100th terms differ by 100, and asked to find how much their 1000th terms differ by.

Solution

Why the gap between two parallel APs never changes

100th term of AP 1 − 100th term of AP 2 = 100 (x + 99d) − (y + 99d) = 100 x − y + 99d − 99d = 100 → x − y = 100 (the 99d cancels out) 1000th term of AP 1 − 1000th term of AP 2: (x + 999d) − (y + 999d) = x − y + 999d − 999d = x − y = 100

✅ The difference between the 1000th terms is also 100 — exactly the same as the 100th terms.

💡 Key insight: If two APs have the same common difference, the gap between their corresponding terms (1st vs 1st, 50th vs 50th, 1000th vs 1000th — any matching pair) is always the same, equal to the difference between their first terms. The "nd" part always cancels out.

Questions 11–13 — Counting Terms in Real Number Ranges

These three questions apply the AP idea to counting problems: numbers divisible by a fixed value, multiples lying in a range, and comparing the position at which two different APs produce the same value. In each case, the first step is recognising the hidden AP.

Q.	Problem	Hidden AP	Answer
11	3-digit numbers divisible by 7	105, 112, ..., 994 (a=105, d=7)	128 numbers
12	Multiples of 4 between 10 and 250	12, 16, ..., 248 (a=12, d=4)	60 multiples
13	n for which nth terms of 63,65,67,... and 3,10,17,... are equal	a=63,d=2 vs a=3,d=7	n = 13

Worked Example — Question 11

How many three-digit numbers are divisible by 7?

Smallest 3-digit multiple of 7 = 105. Largest = 994. This forms an AP: 105, 112, 117, ..., 994 with a=105, d=7 994 = 105 + (n−1)(7) → (n−1) = 889/7 = 127 n = 127 + 1 = 128

Worked Example — Question 13

For what value of n are the nth terms of 63,65,67,... and 3,10,17,... equal?

AP 1: a=63, d=2 → nth term = 63 + (n−1)(2) AP 2: a=3, d=7 → nth term = 3 + (n−1)(7) Setting them equal: 63 + 2(n−1) = 3 + 7(n−1) 63 + 2n − 2 = 3 + 7n − 7 61 + 2n = 7n − 4 → 65 = 5n n = 13

Questions 14 & 16 — Determining an Entire AP from Given Conditions

In both these questions, you're not asked for a single term — you're asked to reconstruct the whole AP, which just means finding a and d, then writing out the first few terms.

Question 14

3rd term is 16; the 7th term exceeds the 5th term by 12

a + 2d = 16 (eq. 1) a₇ = a₅ + 12 → (a+6d) = (a+4d) + 12 → 2d = 12 → d = 6 From (1): a + 2(6) = 16 → a = 16 − 12 = 4

✅ The required AP is 4, 10, 16, 22, ... (a = 4, d = 6)

Question 16

Sum of the 4th and 8th terms is 24; sum of the 6th and 10th terms is 44

a₄+a₈ = 24 → (a+3d)+(a+7d) = 24 → 2a + 10d = 24 (eq. 1) a₆+a₁₀ = 44 → (a+5d)+(a+9d) = 44 → 2a + 14d = 44 (eq. 2) Subtracting (1) from (2): 4d = 20 → d = 5 From (1): 2a + 10(5) = 24 → 2a = −26 → a = −13

✅ The first three terms are −13, −8, −3 (a = −13, d = 5)

Question 15 — Finding a Term Counted From the End

When an AP has a known last term, you can count backwards from the end instead of forwards from the start. The cleanest method: find the total number of terms, then convert "kth from the end" into the equivalent position counted from the beginning.

kth term from the end = (Total terms − k + 1)th term from the start

Solution

Find the 20th term from the end of the AP: 3, 8, 13, ..., 253

a = 3, d = 5. First find the total number of terms (n): 253 = 3 + (n−1)(5) → n−1 = 250/5 = 50 → n = 51 20th from the end = (51 − 20 + 1)th from the start = 32nd term a₃₂ = a + 31d = 3 + 31(5) = 3 + 155 = 158

📌 Alternative method: You can also reverse the AP itself — treat the last term as the new "first term" and the negative of d as the new common difference — then apply the usual nth-term formula directly. Both methods give the same answer.

Question 17 — Real-Life Application: Subba Rao's Salary

Subba Rao started his job in 1995 at a monthly salary of Rs. 5,000, with a fixed yearly increment of Rs. 200. The question asks in which year his salary reached Rs. 7,000 — a classic real-life AP problem, structurally identical to the taxi-fare and well-digging situations from Exercise 6.1.

Solution

In which year did Subba Rao's salary reach Rs. 7,000?

Salaries year-wise: 5000, 5200, 5400, ... — an AP with a = 5000, d = 200 Let the nth term (the year his salary = 7000) be aₙ = 7000: 5000 + (n−1)(200) = 7000 (n−1)(200) = 2000 → n−1 = 10 → n = 11

✅ The 11th term corresponds to the 11th year starting from 1995 (counting 1995 as year 1) — that is, the year 2005, when his salary reached Rs. 7,000.

Common Mistakes to Avoid

Writing (n−1)d as nd: The most frequent slip in this entire exercise — remember it's always (n−1) multiplied by d, not n itself, because the first term needs zero d's added.
Assuming the smaller-numbered given term is a₁: In Question 3 parts (ii) and (v), and in several other questions, a₁ is not given directly and must be calculated — never assume it.
Forgetting that n must be a positive integer: As in Question 6, if solving for n gives a fraction, the value you tested is simply not a term of that AP — there's no rounding allowed.
Sign errors when subtracting equations: Questions 7, 8, 14, and 16 all involve subtracting one linear equation from another — keep careful track of signs, especially when both equations have negative terms.
Miscounting "from the end": In Question 15, remember the formula is (Total terms − k + 1), not simply (Total terms − k) — forgetting the "+1" is an extremely common error.
Off-by-one errors in real-life year/position problems: In Question 17, the year 1995 itself is the first term (n=1), so the salary reaches Rs. 7000 in the 11th year from 1995, which is 2005 — not 1995+11=2006.

⛔ High-value exam tip: Questions 7, 8, 14, and 16 (where two equations must be solved simultaneously for a and d) are extremely popular in board exams as standalone 4-mark questions. Practising the elimination method (subtracting one equation from another to cancel "a") until it's automatic will save valuable time.

Quick Reference — All Answers at a Glance

Question	Topic	Answer
Q1	Fill in the blanks (5 parts)	28, 2, 46, 10, 3.5
Q2	30th & 11th terms	−77, 22
Q3	Missing terms (5 parts)	See table above
Q4	Which term is 78	16th term
Q5	Number of terms (2 parts)	34, 27
Q6	Is −150 a term?	No
Q7	31st term from a₁₁, a₁₆	178
Q8	Which term is zero	5th term
Q9	Common difference	d = 1
Q10	Difference between 1000th terms	100
Q11	3-digit numbers divisible by 7	128
Q12	Multiples of 4 between 10–250	60
Q13	n for equal nth terms	n = 13
Q14	Determine the AP	4, 10, 16, 22, ...
Q15	20th term from the end	158
Q16	First three terms	−13, −8, −3
Q17	Subba Rao's salary reaches 7000	Year 2005

What This Exercise Prepares You For

Mastering the nth-term formula in Exercise 6.2 directly sets up the next big idea in Chapter 6 — the sum of the first n terms of an AP, given by Sₙ = n/2 [2a + (n−1)d]. Many sum-related questions reuse the exact same "find a and d from given conditions" technique practised heavily in Questions 3, 7, 8, 14, and 16 here.

The simultaneous-equation skills used throughout this exercise also connect back to Linear Equations, while the real-life salary problem in Question 17 mirrors the taxi-fare and well-digging situations explored in Exercise 6.1 — revisiting that exercise is excellent preparation before attempting this one.

📐 Board Exam Tip (CBSE, Telangana & AP): Questions involving two unknowns (a and d) solved through simultaneous equations — such as Q7, Q8, Q14, and Q16 — are among the most frequently repeated question types from this exercise in board exams. Make sure you can set up and solve these equations confidently without referring back to examples.