Solved Examples

How to Use This Section

This is your practice powerhouse for Arithmetic Progressions. Below are 30+ fully worked problems spanning the whole chapter — identifying APs, the nth term, sums, arithmetic mean, selection of terms, and word problems — arranged roughly easy to hard.

How to study: Try each problem with the solution covered, then check the steps. Board marks are awarded step by step.

Keep these formulae handy:

• nth term: $a_n = a + (n-1)d$.
• Sum: $S_n = \dfrac{n}{2}[2a + (n-1)d] = \dfrac{n}{2}(a + l)$.
• $a_n = S_n - S_{n-1}$.
• Three terms in AP: $a - d, a, a + d$.
• AM of $a, b$: $\dfrac{a+b}{2}$.

Solved Examples

Example 1: Find $d$

Find the common difference of $11, 8, 5, 2, \dots$

Solution:

1. $d = 8 - 11 = -3$.

Final Answer: $d = -3$.

Takeaway: $d$ = any term minus the previous one.

Example 2: nth term

Find the 20th term of $3, 8, 13, \dots$

Solution:

1. $a = 3$, $d = 5$. $a_{20} = 3 + 19(5) = 98$.

Final Answer: 98.

Takeaway: Apply $a_n = a + (n-1)d$.

Example 3: Which term?

Which term of $5, 9, 13, \dots$ is 81?

Solution:

1. $a = 5$, $d = 4$. $5 + (n-1)4 = 81 \Rightarrow 4(n-1) = 76 \Rightarrow n = 20$.

Final Answer: The 20th term.

Takeaway: Set $a_n$ equal to the value and solve for $n$.

Example 4: Find AP from two terms

The 4th term of an AP is 9 and the 9th term is 24. Find the AP.

Solution:

1. $a + 3d = 9$, $a + 8d = 24$. Subtract: $5d = 15 \Rightarrow d = 3$; $a = 0$.
2. AP: $0, 3, 6, 9, \dots$

Final Answer: $0, 3, 6, 9, \dots$

Takeaway: Two terms ⇒ subtract to get $d$, then $a$.

Example 5: Sum of first n terms

Find the sum of the first 15 terms of $2, 5, 8, \dots$

Solution:

1. $a = 2$, $d = 3$. $S_{15} = \dfrac{15}{2}[4 + 14(3)] = \dfrac{15}{2}(46) = 345$.

Final Answer: 345.

Takeaway: $S_n = \dfrac{n}{2}[2a + (n-1)d]$.

Example 6: Sum with last term

Find the sum $2 + 4 + 6 + \dots + 200$.

Solution:

1. $a = 2$, $l = 200$, $d = 2$. Number of terms: $200 = 2 + (n-1)2 \Rightarrow n = 100$.
2. $S = \dfrac{100}{2}(2 + 200) = 50(202) = 10100$.

Final Answer: 10100.

Takeaway: Find $n$, then use $\dfrac{n}{2}(a + l)$.

Example 7: Arithmetic mean

The AM of two numbers is 17 and one number is 12. Find the other.

Solution:

1. $\dfrac{12 + x}{2} = 17 \Rightarrow x = 22$.

Final Answer: 22.

Takeaway: AM is the average; set up and solve.

Example 8: Three terms in AP

Three numbers in AP have sum 27 and product 504. Find them.

Solution:

1. $a - d, a, a + d$: $3a = 27 \Rightarrow a = 9$.
2. $9(81 - d^2) = 504 \Rightarrow 81 - d^2 = 56 \Rightarrow d^2 = 25 \Rightarrow d = \pm 5$.
3. Numbers: $4, 9, 14$.

Final Answer: 4, 9, 14.

Takeaway: Symmetric terms make the sum give $a$.

Example 9: nth term from end

Find the 6th term from the end of $17, 14, 11, \dots, -40$.

Solution:

1. $l = -40$, $d = -3$. From end: $l - (n-1)d = -40 - 5(-3) = -40 + 15 = -25$.

Final Answer: $-25$.

Takeaway: Use $l - (n-1)d$ from the end.

Example 10: How many terms in an AP

How many terms are there in $7, 13, 19, \dots, 205$?

Solution:

1. $a = 7$, $d = 6$, $l = 205$. $205 = 7 + (n-1)6 \Rightarrow 198 = 6(n-1) \Rightarrow n = 34$.

Final Answer: 34 terms.

Takeaway: Solve $l = a + (n-1)d$ for $n$.

Example 11: Sum of first n natural numbers

Find $1 + 2 + 3 + \dots + 50$.

Solution:

1. $S = \dfrac{50 \times 51}{2} = 1275$.

Final Answer: 1275.

Takeaway: $\dfrac{n(n+1)}{2}$ for the first $n$ naturals.

Example 12: Find $n$ for a sum

How many terms of $18, 16, 14, \dots$ give a sum of 78?

Solution:

1. $a = 18$, $d = -2$. $\dfrac{n}{2}[36 + (n-1)(-2)] = 78 \Rightarrow n(38 - 2n) = 156$.
2. $2n^2 - 38n + 156 = 0 \Rightarrow n^2 - 19n + 78 = 0 \Rightarrow (n-6)(n-13) = 0$.
3. $n = 6$ or $13$.

Final Answer: 6 or 13 terms.

Takeaway: Both positive roots may be valid.

Example 13: Term from sum formula

If $S_n = 2n^2 + 3n$, find the 10th term.

Solution:

1. $a_{10} = S_{10} - S_9$. $S_{10} = 200 + 30 = 230$; $S_9 = 162 + 27 = 189$.
2. $a_{10} = 230 - 189 = 41$.

Final Answer: 41.

Takeaway: $a_n = S_n - S_{n-1}$.

Example 14: Middle term

Find the middle term of the AP $3, 7, 11, \dots, 47$.

Solution:

1. $a = 3$, $d = 4$, $l = 47$. $47 = 3 + (n-1)4 \Rightarrow n = 12$ (even number of terms).
2. With 12 terms there are two middle terms, the 6th and 7th: $a_6 = 3 + 5(4) = 23$, $a_7 = 27$.

Final Answer: Middle terms 23 and 27.

Takeaway: An even count has two middle terms; an odd count has one.

Example 15: Insert means

Insert 4 arithmetic means between 4 and 24.

Solution:

1. $d = \dfrac{24 - 4}{4 + 1} = 4$.
2. Means: $8, 12, 16, 20$.

Final Answer: 8, 12, 16, 20.

Takeaway: $d = \dfrac{b - a}{k + 1}$ for $k$ means.

Example 16: Sum of three-digit multiples of 5

Find the sum of all three-digit numbers divisible by 5.

Solution:

1. AP $100, 105, \dots, 995$. $a = 100$, $d = 5$, $l = 995$. $995 = 100 + (n-1)5 \Rightarrow n = 180$.
2. $S = \dfrac{180}{2}(100 + 995) = 90(1095) = 98550$.

Final Answer: 98550.

Takeaway: Find $n$ first, then the sum with first + last.

Example 17: First negative term

Find the first negative term of $50, 47, 44, \dots$

Solution:

1. $a = 50$, $d = -3$. $a_n < 0$: $50 + (n-1)(-3) < 0 \Rightarrow 53 - 3n < 0 \Rightarrow n > 17.67$.
2. First integer $n = 18$: $a_{18} = 50 + 17(-3) = -1$.

Final Answer: 18th term, $-1$.

Takeaway: Solve the inequality $a_n < 0$ for the smallest integer $n$.

Example 18: Sum of a part of an AP

Find the sum of the terms from the 5th to the 12th of $2, 5, 8, \dots$

Solution:

1. $a = 2$, $d = 3$. $S_{12} - S_4$. $S_{12} = \dfrac{12}{2}[4 + 33] = 6(37) = 222$; $S_4 = \dfrac{4}{2}[4 + 9] = 2(13) = 26$.
2. Required $= 222 - 26 = 196$.

Final Answer: 196.

Takeaway: Sum from term $p$ to $q$ is $S_q - S_{p-1}$.

Example 19: Equation in $n$

For what value of $n$ are the $n$th terms of $63, 65, 67, \dots$ and $3, 10, 17, \dots$ equal?

Solution:

1. First AP: $a_n = 63 + (n-1)2 = 61 + 2n$. Second: $a_n = 3 + (n-1)7 = 7n - 4$.
2. Set equal: $61 + 2n = 7n - 4 \Rightarrow 65 = 5n \Rightarrow n = 13$.

Final Answer: $n = 13$.

Takeaway: Write each nth term and equate.

Example 20: Sum given $a$ and $l$

The first term of an AP is 5, the last term is 45, and the sum is 400. Find the number of terms.

Solution:

1. $S_n = \dfrac{n}{2}(a + l) = \dfrac{n}{2}(5 + 45) = 25n = 400 \Rightarrow n = 16$.

Final Answer: 16 terms.

Takeaway: With $a$ and $l$ known, $S_n = \dfrac{n}{2}(a+l)$ gives $n$ directly.

Example 21: Find $d$ then a later sum

The first term of an AP is 5, the last term 45, sum 400. Find the common difference.

Solution:

1. From Example 20, $n = 16$. Then $l = a + (n-1)d$: $45 = 5 + 15d \Rightarrow d = \dfrac{40}{15} = \dfrac{8}{3}$.

Final Answer: $d = \dfrac{8}{3}$.

Takeaway: Use $l = a + (n-1)d$ once $n$ is known.

Example 22: Given sum of two terms

The sum of the 4th and 8th terms of an AP is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms.

Solution:

1. $a_4 + a_8 = (a+3d)+(a+7d) = 2a + 10d = 24$.
2. $a_6 + a_{10} = (a+5d)+(a+9d) = 2a + 14d = 44$.
3. Subtract: $4d = 20 \Rightarrow d = 5$; then $2a + 50 = 24 \Rightarrow a = -13$.
4. First three: $-13, -8, -3$.

Final Answer: $-13, -8, -3$.

Takeaway: Form two equations from the term-sum conditions.

Example 23: Sum of even numbers

Find the sum of the first 40 positive even numbers.

Solution:

1. AP $2, 4, 6, \dots$: $a = 2$, $d = 2$, $n = 40$.
2. $S_{40} = \dfrac{40}{2}[4 + 39(2)] = 20[4 + 78] = 20(82) = 1640$.

Final Answer: 1640.

Takeaway: Sum of first $n$ even numbers $= n(n+1)$ (here $40 \times 41 = 1640$).

Example 24: Word problem — savings target

Kavita saves ₹50 in week 1, increasing ₹10 each week. In how many weeks will her total savings reach ₹1100?

Solution:

1. $a = 50$, $d = 10$. $\dfrac{n}{2}[100 + (n-1)10] = 1100 \Rightarrow n(90 + 10n) = 2200$.
2. $10n^2 + 90n - 2200 = 0 \Rightarrow n^2 + 9n - 220 = 0 \Rightarrow (n + 20)(n - 11) = 0$.
3. $n = 11$ (reject $-20$).

Final Answer: 11 weeks.

Takeaway: Set $S_n$ equal to the target and solve the quadratic.

Example 25: Ratio of terms

The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th to the 21st term.

Solution:

1. $\dfrac{a + 10d}{a + 17d} = \dfrac{2}{3} \Rightarrow 3a + 30d = 2a + 34d \Rightarrow a = 4d$.
2. $\dfrac{a + 4d}{a + 20d} = \dfrac{4d + 4d}{4d + 20d} = \dfrac{8d}{24d} = \dfrac{1}{3}$.

Final Answer: 1 : 3.

Takeaway: Express $a$ in terms of $d$ from the given ratio, then substitute.

Example 26: Sum of multiples between bounds

Find the sum of all multiples of 4 lying between 10 and 250.

Solution:

1. Multiples: $12, 16, \dots, 248$. $a = 12$, $d = 4$, $l = 248$. $248 = 12 + (n-1)4 \Rightarrow n = 60$.
2. $S = \dfrac{60}{2}(12 + 248) = 30(260) = 7800$.

Final Answer: 7800.

Takeaway: Identify the first and last multiples within the range.

Example 27: nth term given $S_n$ pattern

The sum of $n$ terms of an AP is $S_n = n^2$. Show it is an AP and find $d$.

Solution:

1. $a_n = S_n - S_{n-1} = n^2 - (n-1)^2 = 2n - 1$.
2. $a_1 = 1$, $a_2 = 3$, so $d = 2$ (the odd numbers).

Final Answer: AP of odd numbers, $d = 2$.

Takeaway: $S_n = n^2$ corresponds to $1 + 3 + 5 + \dots$

Example 28: Which term is double another

In an AP with $a = 2$, $d = 3$, which term is double the 5th term?

Solution:

1. $a_5 = 2 + 4(3) = 14$. We want $a_n = 28$.
2. $2 + (n-1)3 = 28 \Rightarrow 3(n-1) = 26 \Rightarrow n - 1 = \dfrac{26}{3}$, not an integer.
3. So no term is exactly double the 5th term.

Final Answer: No such term exists.

Takeaway: A non-integer $n$ means the required term is not in the AP.

Example 29: Sum equals a term position

The 14th term of an AP is twice its 8th term. If the 6th term is $-8$, find the sum of the first 20 terms.

Solution:

1. $a_{14} = 2a_8$: $a + 13d = 2(a + 7d) \Rightarrow -a - d = 0 \Rightarrow a = -d$.
2. $a_6 = a + 5d = -8$. With $a = -d$: $-d + 5d = -8 \Rightarrow 4d = -8 \Rightarrow d = -2$, $a = 2$.
3. $S_{20} = \dfrac{20}{2}[4 + 19(-2)] = 10[4 - 38] = 10(-34) = -340$.

Final Answer: $-340$.

Takeaway: Translate the term relations into equations for $a$ and $d$ first.

Example 30: Reverse AP sum

Find the sum of the AP $25, 22, 19, \dots, 1$.

Solution:

1. $a = 25$, $d = -3$, $l = 1$. $1 = 25 + (n-1)(-3) \Rightarrow 24 = 3(n-1) \Rightarrow n = 9$.
2. $S = \dfrac{9}{2}(25 + 1) = \dfrac{9}{2}(26) = 117$.

Final Answer: 117.

Takeaway: Even decreasing APs use the same formulas.

Example 31: Mixed multi-step

The sum of the first 7 terms of an AP is 49 and that of the first 17 terms is 289. Find the sum of the first $n$ terms.

Solution:

1. $S_7 = \dfrac{7}{2}(2a + 6d) = 7(a + 3d) = 49 \Rightarrow a + 3d = 7$.
2. $S_{17} = \dfrac{17}{2}(2a + 16d) = 17(a + 8d) = 289 \Rightarrow a + 8d = 17$.
3. Subtract: $5d = 10 \Rightarrow d = 2$; $a = 1$.
4. $S_n = \dfrac{n}{2}[2 + (n-1)2] = \dfrac{n}{2}(2n) = n^2$.

Final Answer: $S_n = n^2$.

Takeaway: Two sum-conditions give $a$ and $d$, then the general $S_n$.