CBSE 2026-27 Syllabus Scope Note:
Arithmetic Progressions is fully retained in the curriculum: identifying APs & common differences, finding the general nth term, calculating the sum of first n terms, and solving real-world day-to-day contextual problems. Every problem from Exercises 5.1, 5.2, 5.3, and 5.4 is solved below.
Exercise 5.1: Multiple Choice Questions (MCQs)
Exercise 5.1 — Q1
1. In an AP, if \(d = -4\), \(n = 7\), \(a_n = 4\), then \(a\) is:
(A) \(6\)
(B) \(7\)
(C) \(20\)
(D) \(28\)
Solution: Correct Option (D)
Using the general formula \(a_n = a + (n – 1)d\):
\[4 = a + (7 – 1)(-4) \implies 4 = a – 24 \implies a = 28\]
Exercise 5.1 — Q2
2. In an AP, if \(a = 3.5\), \(d = 0\), \(n = 101\), then \(a_n\) will be:
(A) \(0\)
(B) \(3.5\)
(C) \(103.5\)
(D) \(104.5\)
Solution: Correct Option (B)
\[a_{101} = a + (101 – 1)d = 3.5 + 100(0) = 3.5\]
Exercise 5.1 — Q3
3. The list of numbers \(-10, -6, -2, 2, \dots\) is:
(A) an AP with \(d = -16\)
(B) an AP with \(d = 4\)
(C) an AP with \(d = -4\)
(D) not an AP
Solution: Correct Option (B)
Differences between consecutive terms:
\[-6 – (-10) = 4, \quad -2 – (-6) = 4, \quad 2 – (-2) = 4\]
Since the difference is constant, it is an AP with \(d = 4\).
Exercise 5.1 — Q4
4. The \(11^{\text{th}}\) term of the AP: \(-5, -\frac{5}{2}, 0, \frac{5}{2}, \dots\) is:
(A) \(-20\)
(B) \(20\)
(C) \(-30\)
(D) \(30\)
Solution: Correct Option (B)
Here \(a = -5\) and \(d = -\frac{5}{2} – (-5) = \frac{5}{2}\):
\[a_{11} = -5 + (11 – 1)\left(\frac{5}{2}\right) = -5 + 10\left(\frac{5}{2}\right) = -5 + 25 = 20\]
Exercise 5.1 — Q5
5. The first four terms of an AP, whose first term is \(-2\) and common difference is \(-2\), are:
(A) \(-2, 0, 2, 4\)
(B) \(-2, -4, -8, -16\)
(C) \(-2, -4, -6, -8\)
(D) \(-2, -4, -8, -16\)
Solution: Correct Option (C)
\[a_1 = -2, \quad a_2 = -2 + (-2) = -4, \quad a_3 = -4 + (-2) = -6, \quad a_4 = -6 + (-2) = -8\]
Exercise 5.1 — Q6
6. The \(21^{\text{st}}\) term of the AP whose first two terms are \(-3\) and \(4\) is:
(A) \(17\)
(B) \(137\)
(C) \(143\)
(D) \(-143\)
Solution: Correct Option (B)
\(a = -3\), \(d = 4 – (-3) = 7\):
\[a_{21} = -3 + 20(7) = -3 + 140 = 137\]
Exercise 5.1 — Q7
7. If the \(2^{\text{nd}}\) term of an AP is \(13\) and the \(5^{\text{th}}\) term is \(25\), what is its \(7^{\text{th}}\) term?
(A) \(30\)
(B) \(33\)
(C) \(37\)
(D) \(38\)
Solution: Correct Option (B)
\(a_2 = a + d = 13\) and \(a_5 = a + 4d = 25\). Subtracting yields \(3d = 12 \implies d = 4\). Then \(a = 9\):
\[a_7 = 9 + 6(4) = 33\]
Exercise 5.1 — Q8
8. Which term of the AP: \(21, 42, 63, 84, \dots\) is \(210\)?
(A) \(9^{\text{th}}\)
(B) \(10^{\text{th}}\)
(C) \(11^{\text{th}}\)
(D) \(12^{\text{th}}\)
Solution: Correct Option (B)
\(a = 21\), \(d = 21\):
\[210 = 21 + (n – 1)21 \implies 21n = 210 \implies n = 10\]
Exercise 5.1 — Q9
9. If the common difference of an AP is \(5\), then what is \(a_{18} – a_{13}\)?
(A) \(5\)
(B) \(20\)
(C) \(25\)
(D) \(30\)
Solution: Correct Option (C)
\[a_{18} – a_{13} = (a + 17d) – (a + 12d) = 5d = 5(5) = 25\]
Exercise 5.1 — Q10
10. What is the common difference of an AP in which \(a_{18} – a_{14} = 32\)?
(A) \(8\)
(B) \(-8\)
(C) \(-4\)
(D) \(4\)
Solution: Correct Option (A)
\[a_{18} – a_{14} = (a + 17d) – (a + 13d) = 4d = 32 \implies d = 8\]
Exercise 5.1 — Q11
11. Two APs have the same common difference. The first term of one is \(-1\) and that of the other is \(-8\). The difference between their \(4^{\text{th}}\) terms is:
(A) \(-1\)
(B) \(-8\)
(C) \(7\)
(D) \(-9\)
Solution: Correct Option (C)
\[a_4 – b_4 = (a + 3d) – (b + 3d) = a – b = -1 – (-8) = 7\]
Exercise 5.1 — Q12
12. If \(7\) times the \(7^{\text{th}}\) term of an AP is equal to \(11\) times its \(11^{\text{th}}\) term, then its \(18^{\text{th}}\) term will be:
(A) \(7\)
(B) \(11\)
(C) \(18\)
(D) \(0\)
Solution: Correct Option (D)
\[7(a + 6d) = 11(a + 10d) \implies 7a + 42d = 11a + 110d \implies 4a + 68d = 0 \implies a + 17d = 0\]
\[a_{18} = a + 17d = 0\]
Exercise 5.1 — Q13
13. The \(4^{\text{th}}\) term from the end of the AP: \(-11, -8, -5, \dots, 49\) is:
(A) \(37\)
(B) \(40\)
(C) \(43\)
(D) \(58\)
Solution: Correct Option (B)
\(l = 49\), \(d = 3\):
\[l – (4 – 1)d = 49 – 3(3) = 40\]
Exercise 5.1 — Q14
14. The famous mathematician associated with finding the sum of the first 100 natural numbers is:
(A) Pythagoras
(B) Newton
(C) Gauss
(D) Euclid
Solution: Correct Option (C)
Carl Friedrich Gauss famously deduced the formula for the sum of consecutive natural numbers as a young student.
Exercise 5.1 — Q15
15. If the first term of an AP is \(-5\) and the common difference is \(2\), then the sum of the first \(6\) terms is:
(A) \(0\)
(B) \(5\)
(C) \(6\)
(D) \(15\)
Solution: Correct Option (A)
\[S_6 = \frac{6}{2}[2(-5) + (6 – 1)(2)] = 3[-10 + 10] = 0\]
Exercise 5.1 — Q16
16. The sum of first \(16\) terms of the AP: \(10, 6, 2, \dots\) is:
(A) \(-320\)
(B) \(320\)
(C) \(-352\)
(D) \(-400\)
Solution: Correct Option (A)
\[S_{16} = \frac{16}{2}[2(10) + 15(-4)] = 8[20 – 60] = -320\]
Exercise 5.1 — Q17
17. In an AP if \(a = 1\), \(a_n = 20\) and \(S_n = 399\), then \(n\) is:
(A) \(19\)
(B) \(21\)
(C) \(38\)
(D) \(42\)
Solution: Correct Option (C)
\[399 = \frac{n}{2}(1 + 20) \implies 399 = \frac{21n}{2} \implies n = 38\]
Exercise 5.1 — Q18
18. The sum of first five multiples of \(3\) is:
(A) \(45\)
(B) \(55\)
(C) \(65\)
(D) \(75\)
Solution: Correct Option (A)
The sequence is \(3, 6, 9, 12, 15\):
\[S_5 = \frac{5}{2}(3 + 15) = \frac{5}{2}(18) = 45\]
Exercise 5.2: Short Answer Questions with Reasoning
Exercise 5.2 — Q1
1. Which of the following form an AP? Justify your answer.
(i) \(-1, -1, -1, -1, \dots\)
(ii) \(0, 2, 0, 2, \dots\)
(iii) \(1, 1, 2, 2, 3, 3, \dots\)
(iv) \(11, 22, 33, \dots\)
(v) \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots\)
(vi) \(2, 2^2, 2^3, 2^4, \dots\)
(vii) \(\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \dots\)
Solution
(i) Yes: \(d = -1 – (-1) = 0\) (constant difference).
(ii) No: \(a_2 – a_1 = 2 – 0 = 2\), but \(a_3 – a_2 = 0 – 2 = -2 \neq 2\).
(iii) No: \(a_2 – a_1 = 1 – 1 = 0\), but \(a_3 – a_2 = 2 – 1 = 1 \neq 0\).
(iv) Yes: \(22 – 11 = 11, \, 33 – 22 = 11 \implies d = 11\).
(v) No: \(\frac{1}{3} – \frac{1}{2} = -\frac{1}{6}\), while \(\frac{1}{4} – \frac{1}{3} = -\frac{1}{12} \neq -\frac{1}{6}\).
(vi) No: \(4 – 2 = 2\), while \(8 – 4 = 4 \neq 2\).
(vii) Yes: \(\sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3} \implies d = \sqrt{3}\) (constant).
Exercise 5.2 — Q2
2. Justify whether it is true to say that \(-1, -\frac{3}{2}, -2, -\frac{5}{2}, \dots\) forms an AP as \(a_2 – a_1 = a_3 – a_2\).
Solution: Yes
\[a_2 – a_1 = -\frac{3}{2} – (-1) = -\frac{1}{2}\]
\[a_3 – a_2 = -2 – \left(-\frac{3}{2}\right) = -\frac{1}{2}\]
\[a_4 – a_3 = -\frac{5}{2} – (-2) = -\frac{1}{2}\]
Since \(a_{k+1} – a_k = -\frac{1}{2}\) for all consecutive terms, the statement is true.
Exercise 5.2 — Q3
3. For the AP: \(-3, -7, -11, \dots\), can we find directly \(a_{30} – a_{20}\) without actually finding \(a_{30}\) and \(a_{20}\)? Give reasons.
Solution: Yes
\(a_{30} – a_{20} = [a + 29d] – [a + 19d] = 10d\). Here \(d = -7 – (-3) = -4\):
\[a_{30} – a_{20} = 10(-4) = -40\]
Exercise 5.2 — Q4
4. Two APs have the same common difference. The first term of one AP is \(2\) and that of the other is \(7\). The difference between their \(10^{\text{th}}\) terms is the same as the difference between their \(21^{\text{st}}\) terms. Why?
Solution
Let the \(n^{\text{th}}\) terms be \(a_n = 2 + (n – 1)d\) and \(b_n = 7 + (n – 1)d\):
\[b_n – a_n = [7 + (n – 1)d] – [2 + (n – 1)d] = 5\]
Since the difference is completely independent of \(n\), it remains identical for all corresponding terms.
Exercise 5.2 — Q5
5. Is \(0\) a term of the AP: \(31, 28, 25, \dots\)? Justify your answer.
Solution: No
\[31 + (n – 1)(-3) = 0 \implies 3(n – 1) = 31 \implies n = \frac{34}{3} = 11\frac{1}{3}\]
Since \(n\) must be a positive integer, \(0\) cannot be a term of this AP.
Exercise 5.2 — Q6
6. The taxi fare after each km, when fare is \(\text{Rs } 15\) for first km and \(\text{Rs } 8\) for each additional km, does not form an AP as the total fare after each km is \(15, 8, 8, 8, \dots\). Is this statement true?
Solution: False
The total fare after each km is: \(1^{\text{st}}\text{ km} = 15\), \(2^{\text{nd}}\text{ km} = 15 + 8 = 23\), \(3^{\text{rd}}\text{ km} = 23 + 8 = 31\), giving the list \(15, 23, 31, 39, \dots\), which does form an AP with \(a = 15\) and \(d = 8\).
Exercise 5.2 — Q7
7. In which of the following situations do the lists of numbers form an AP?
(i) School monthly fee of \(\text{Rs } 400\) for the whole session.
(ii) School fee starting at \(\text{Rs } 250\) for Class I, increasing by \(\text{Rs } 50\) for each higher class.
(iii) Amount in account at simple interest of \(10\%\) p.a. on \(\text{Rs } 1000\).
(iv) Bacteria count doubling every second.
Solution
(i) Yes: \(400, 400, 400, \dots\) has common difference \(d = 0\).
(ii) Yes: \(250, 300, 350, \dots\) has constant common difference \(d = 50\).
(iii) Yes: Annual simple interest is constant (\(\text{Rs } 100\)), so amounts \(1100, 1200, 1300, \dots\) have \(d = 100\).
(iv) No: Numbers are \(x, 2x, 4x, 8x, \dots\); differences \(x, 2x, 4x\) are not constant.
Exercise 5.2 — Q8
8. Justify whether the following can be the \(n^{\text{th}}\) terms of an AP:
(i) \(2n – 3\) (ii) \(3n^2 + 5\) (iii) \(1 + n + n^2\)
Solution
(i) Yes: \(a_n = 2n – 3\) is a linear polynomial in \(n\). \(a_n – a_{n-1} = 2\) is constant.
(ii) No: Degree 2 (quadratic), so consecutive differences depend on \(n\).
(iii) No: Degree 2 polynomial in \(n\), not an AP.
Exercise 5.3: Short Answer Questions
Exercise 5.3 — Q1
1. Match Column A with suitable common differences in Column B:
Solution
• \((A_1)\; 2, -2, -6, -10, \dots \implies d = -2 – 2 = -4 \to \mathbf{(B_5)}\)
• \((A_2)\; a = -18, n = 10, a_n = 0 \implies 0 = -18 + 9d \implies d = 2 \to \mathbf{(B_7)}\)
• \((A_3)\; a = 0, a_{10} = 6 \implies 6 = 0 + 9d \implies d = \frac{2}{3} \to \mathbf{(B_1)}\)
• \((A_4)\; a_2 = 13, a_4 = 3 \implies 2d = 3 – 13 = -10 \implies d = -5 \to \mathbf{(B_2)}\)
Exercise 5.3 — Q2
2. Verify that each is an AP, then write its next three terms:
Solution
(i) \(0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \dots\): \(d = \frac{1}{4}\). Next three terms: \(\mathbf{1, \frac{5}{4}, \frac{3}{2}}\).
(ii) \(5, \frac{14}{3}, \frac{13}{3}, 4, \dots\): \(d = -\frac{1}{3}\). Next three terms: \(\mathbf{\frac{11}{3}, \frac{10}{3}, 3}\).
(iii) \(\sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, \dots\): \(d = \sqrt{3}\). Next three terms: \(\mathbf{4\sqrt{3}, 5\sqrt{3}, 6\sqrt{3}}\).
(iv) \((a + b), (a + 1) + b, (a + 1) + (b + 1), \dots\): \(d = 1\). Next three terms: \(\mathbf{(a + 2) + (b + 1), (a + 2) + (b + 2), (a + 3) + (b + 2)}\).
(v) \(a, 2a + 1, 3a + 2, 4a + 3, \dots\): \(d = a + 1\). Next three terms: \(\mathbf{5a + 4, 6a + 5, 7a + 6}\).
Exercise 5.3 — Q3
3. Write the first three terms of the APs with given \(a\) and \(d\):
Solution
(i) \(a = \frac{1}{2}, d = -\frac{1}{6}\): \(\mathbf{\frac{1}{2}, \frac{1}{3}, \frac{1}{6}}\)
(ii) \(a = -5, d = -3\): \(\mathbf{-5, -8, -11}\)
(iii) \(a = \sqrt{2}, d = \frac{1}{\sqrt{2}}\): \(\mathbf{\sqrt{2}, \frac{3\sqrt{2}}{2}, 2\sqrt{2}}\)
Exercise 5.3 — Q4
4. Find \(a, b,\) and \(c\) such that the numbers are in AP: \(a, 7, b, 23, c\).
Solution
\(a_4 – a_2 = 2d \implies 23 – 7 = 16 \implies d = 8\). Then:
\[a = 7 – 8 = -1, \quad b = 7 + 8 = 15, \quad c = 23 + 8 = 31\]
Exercise 5.3 — Q5
5. Determine the AP whose fifth term is \(19\) and the difference of the eighth term from the thirteenth term is \(20\).
Solution
\(a_{13} – a_8 = 5d = 20 \implies d = 4\).
\(a_5 = a + 4(4) = 19 \implies a = 3\).
AP: \(\mathbf{3, 7, 11, 15, \dots}\)
Exercise 5.3 — Q6
6. The \(26^{\text{th}}\), \(11^{\text{th}}\) and last term of an AP are \(0, 3,\) and \(-\frac{1}{5}\) respectively. Find the common difference and the number of terms.
Solution
\[a_{26} – a_{11} = 15d = 0 – 3 = -3 \implies d = -\frac{1}{5}\]
\[a_{26} = a + 25\left(-\frac{1}{5}\right) = 0 \implies a = 5\]
\[a_n = 5 + (n – 1)\left(-\frac{1}{5}\right) = -\frac{1}{5} \implies n – 1 = 26 \implies n = 27\]
Answer: \(d = -\frac{1}{5}, \, n = 27\).
Exercise 5.3 — Q7
7. The sum of the \(5^{\text{th}}\) and \(7^{\text{th}}\) terms of an AP is \(52\) and the \(10^{\text{th}}\) term is \(46\). Find the AP.
Solution
\[(a + 4d) + (a + 6d) = 52 \implies 2a + 10d = 52 \implies a + 5d = 26\]
\[a_{10} = a + 9d = 46\]
Subtracting yields \(4d = 20 \implies d = 5\). Then \(a = 26 – 25 = 1\).
AP: \(\mathbf{1, 6, 11, 16, \dots}\)
Exercise 5.3 — Q8
8. Find the \(20^{\text{th}}\) term of the AP whose \(7^{\text{th}}\) term is \(24\) less than the \(11^{\text{th}}\) term, first term being \(12\).
Solution
\[a_{11} – a_7 = 4d = 24 \implies d = 6\]
\[a_{20} = a + 19d = 12 + 19(6) = 12 + 114 = 126\]
Exercise 5.3 — Q9
9. If the \(9^{\text{th}}\) term of an AP is zero, prove that its \(29^{\text{th}}\) term is twice its \(19^{\text{th}}\) term.
Proof
\(a_9 = a + 8d = 0 \implies a = -8d\).
\[a_{19} = a + 18d = -8d + 18d = 10d\]
\[a_{29} = a + 28d = -8d + 28d = 20d = 2(10d) = 2 \cdot a_{19}\]
Exercise 5.3 — Q10
10. Find whether \(55\) is a term of the AP: \(7, 10, 13, \dots\) or not. If yes, find which term it is.
Solution: Yes
\[7 + (n – 1)3 = 55 \implies 3(n – 1) = 48 \implies n – 1 = 16 \implies n = 17\]
\(55\) is the \(17^{\text{th}}\) term.
Exercise 5.3 — Q11
11. Determine \(k\) so that \(k^2 + 4k + 8\), \(2k^2 + 3k + 6\), \(3k^2 + 4k + 4\) are three consecutive terms of an AP.
Solution
\[2(2k^2 + 3k + 6) = (k^2 + 4k + 8) + (3k^2 + 4k + 4)\]
\[4k^2 + 6k + 12 = 4k^2 + 8k + 12 \implies 2k = 0 \implies k = 0\]
Exercise 5.3 — Q12
12. Split \(207\) into three parts such that they are in AP and the product of the two smaller parts is \(4623\).
Solution
Let parts be \(a – d, a, a + d\):
\[3a = 207 \implies a = 69\]
\[(a – d)(a) = 4623 \implies (69 – d)(69) = 4623 \implies 69 – d = 67 \implies d = 2\]
Parts: \(\mathbf{67, 69, 71}\).
Exercise 5.3 — Q13
13. The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.
Solution
Let angles be \(a – d, a, a + d\). Sum \(= 180^\circ \implies 3a = 180^\circ \implies a = 60^\circ\).
\[a + d = 2(a – d) \implies 60 + d = 120 – 2d \implies 3d = 60 \implies d = 20^\circ\]
Angles: \(\mathbf{40^\circ, 60^\circ, 80^\circ}\).
Exercise 5.3 — Q14
14. If the \(n^{\text{th}}\) terms of two APs: \(9, 7, 5, \dots\) and \(24, 21, 18, \dots\) are the same, find \(n\) and that term.
Solution
\[9 + (n – 1)(-2) = 24 + (n – 1)(-3)\]
\[11 – 2n = 27 – 3n \implies n = 16\]
\[\text{Term } a_{16} = 11 – 2(16) = -21\]
Exercise 5.3 — Q15
15. If the sum of the \(3^{\text{rd}}\) and \(8^{\text{th}}\) terms of an AP is \(7\) and the sum of the \(7^{\text{th}}\) and \(14^{\text{th}}\) terms is \(-3\), find the \(10^{\text{th}}\) term.
Solution
\[(a + 2d) + (a + 7d) = 7 \implies 2a + 9d = 7\]
\[(a + 6d) + (a + 13d) = -3 \implies 2a + 19d = -3\]
Subtracting yields \(10d = -10 \implies d = -1\). Then \(2a = 16 \implies a = 8\).
\[a_{10} = a + 9d = 8 + 9(-1) = -1\]
Exercise 5.3 — Q16
16. Find the \(12^{\text{th}}\) term from the end of the AP: \(-2, -4, -6, \dots, -100\).
Solution
\[l – (12 – 1)d = -100 – 11(-2) = -100 + 22 = -78\]
Exercise 5.3 — Q17
17. Which term of the AP: \(53, 48, 43, \dots\) is the first negative term?
Solution
\[53 + (n – 1)(-5) < 0 \implies 58 < 5n \implies n > 11.6 \implies n = 12\]
The \(12^{\text{th}}\) term is the first negative term.
Exercise 5.3 — Q18
18. How many numbers lie between \(10\) and \(300\), which when divided by \(4\) leave a remainder \(3\)?
Solution
Numbers of form \(4k + 3\): First number is \(11\) (\(4 \times 2 + 3\)), last number is \(299\) (\(4 \times 74 + 3\)):
\[299 = 11 + (n – 1)4 \implies 288 = 4(n – 1) \implies n – 1 = 72 \implies n = 73\]
Exercise 5.3 — Q19
19. Find the sum of the two middle-most terms of the AP: \(-\frac{4}{3}, -1, -\frac{2}{3}, \dots, 4\frac{1}{3}\).
Solution
\(a = -\frac{4}{3}\), \(d = -1 – (-\frac{4}{3}) = \frac{1}{3}\), \(l = \frac{13}{3}\):
\[\frac{13}{3} = -\frac{4}{3} + (n – 1)\frac{1}{3} \implies n – 1 = 17 \implies n = 18\]
Middle terms are \(9^{\text{th}}\) and \(10^{\text{th}}\):
\[a_9 + a_{10} = (a + 8d) + (a + 9d) = 2a + 17d = 2\left(-\frac{4}{3}\right) + 17\left(\frac{1}{3}\right) = \frac{9}{3} = 3\]
Exercise 5.3 — Q20
20. The first term of an AP is \(-5\) and last term is \(45\). If the sum of terms is \(120\), find the number of terms and common difference.
Solution
\[S_n = \frac{n}{2}(-5 + 45) = 120 \implies 20n = 120 \implies n = 6\]
\[45 = -5 + (6 – 1)d \implies 5d = 50 \implies d = 10\]
Exercise 5.3 — Q21
21. Find the sum:
(i) \(1 + (-2) + (-5) + (-8) + \dots + (-236)\)
(ii) \(\left(4 – \frac{1}{n}\right) + \left(4 – \frac{2}{n}\right) + \left(4 – \frac{3}{n}\right) + \dots\) up to \(n\) terms
(iii) \(\frac{a – b}{a + b} + \frac{3a – 2b}{a + b} + \frac{5a – 3b}{a + b} + \dots\) to \(11\) terms
Solution
(i): \(a = 1, d = -3, a_n = -236 \implies -236 = 1 + (n – 1)(-3) \implies n = 80\).
\[S_{80} = \frac{80}{2}(1 – 236) = 40(-235) = -9400\]
(ii): Grouping integer and fractional parts:
\[(4 + 4 + \dots \text{ to } n \text{ terms}) – \frac{1}{n}(1 + 2 + 3 + \dots + n) = 4n – \frac{1}{n} \cdot \frac{n(n + 1)}{2} = \frac{7n – 1}{2}\]
(iii): Common denominator \((a + b)\). First term \(A = \frac{a – b}{a + b}\), \(D = \frac{2a – b}{a + b}\):
\[S_{11} = \frac{11}{2}\left[2\left(\frac{a – b}{a + b}\right) + 10\left(\frac{2a – b}{a + b}\right)\right] = \frac{11(22a – 12b)}{2(a + b)} = \frac{11(11a – 6b)}{a + b}\]
Exercise 5.3 — Q22
22. Which term of the AP: \(-2, -7, -12, \dots\) will be \(-77\)? Find the sum of this AP up to \(-77\).
Solution
\[-2 + (n – 1)(-5) = -77 \implies -5(n – 1) = -75 \implies n = 16\]
\[S_{16} = \frac{16}{2}[-2 + (-77)] = 8(-79) = -632\]
Exercise 5.3 — Q23
23. If \(a_n = 3 – 4n\), show that \(a_1, a_2, a_3, \dots\) form an AP. Also find \(S_{20}\).
Solution
\(a_n – a_{n-1} = (3 – 4n) – [3 – 4(n – 1)] = -4\) (constant difference). Thus it is an AP with \(a = -1, d = -4\).
\[S_{20} = \frac{20}{2}[2(-1) + 19(-4)] = 10[-2 – 76] = -780\]
Exercise 5.3 — Q24
24. In an AP, if \(S_n = n(4n + 1)\), find the AP.
Solution
\(a_1 = S_1 = 1(5) = 5\). \(S_2 = 2(9) = 18 \implies a_2 = S_2 – S_1 = 18 – 5 = 13\).
Common difference \(d = 13 – 5 = 8\). AP: \(\mathbf{5, 13, 21, 29, \dots}\)
Exercise 5.3 — Q25
25. In an AP, if \(S_n = 3n^2 + 5n\) and \(a_k = 164\), find the value of \(k\).
Solution
\[a_n = S_n – S_{n-1} = (3n^2 + 5n) – [3(n-1)^2 + 5(n-1)] = 6n + 2\]
\[6k + 2 = 164 \implies 6k = 162 \implies k = 27\]
Exercise 5.3 — Q26
26. If \(S_n\) denotes the sum of first \(n\) terms of an AP, prove that \(S_{12} = 3(S_8 – S_4)\).
Proof
\[3(S_8 – S_4) = 3(4[2a + 7d] – 2[2a + 3d]) = 3(8a + 28d – 4a – 6d) = 3(4a + 22d) = 12a + 66d\]
\[S_{12} = 6(2a + 11d) = 12a + 66d\]
Exercise 5.3 — Q27
27. Find the sum of first \(17\) terms of an AP whose \(4^{\text{th}}\) and \(9^{\text{th}}\) terms are \(-15\) and \(-30\) respectively.
Solution
\[a_9 – a_4 = 5d = -30 – (-15) = -15 \implies d = -3\]
\[a = -15 – 3(-3) = -6\]
\[S_{17} = \frac{17}{2}[2(-6) + 16(-3)] = \frac{17}{2}[-12 – 48] = \frac{17}{2}(-60) = -510\]
Exercise 5.3 — Q28
28. If the sum of first \(6\) terms is \(36\) and that of first \(16\) terms is \(256\), find the sum of first \(10\) terms.
Solution
\[\frac{6}{2}[2a + 5d] = 36 \implies 2a + 5d = 12\]
\[\frac{16}{2}[2a + 15d] = 256 \implies 2a + 15d = 32\]
Subtracting gives \(10d = 20 \implies d = 2\) and \(a = 1\).
\[S_{10} = \frac{10}{2}[2(1) + 9(2)] = 5(20) = 100\]
Exercise 5.3 — Q29
29. Find the sum of all the \(11\) terms of an AP whose middle-most term is \(30\).
Solution
For \(11\) terms, the middle term is the \(6^{\text{th}}\) term (\(a_6 = a + 5d = 30\)):
\[S_{11} = \frac{11}{2}[2a + 10d] = 11(a + 5d) = 11(30) = 330\]
Exercise 5.3 — Q30
30. Find the sum of last ten terms of the AP: \(8, 10, 12, \dots, 126\).
Solution
Reversing the AP with last term as first: \(a = 126, d = -2, n = 10\):
\[S = \frac{10}{2}[2(126) + 9(-2)] = 5[252 – 18] = 5(234) = 1170\]
Exercise 5.3 — Q31
31. Find the sum of first seven numbers which are multiples of \(2\) as well as of \(9\).
Solution
\(\text{LCM}(2, 9) = 18\). The sequence is \(18, 36, 54, \dots\) to \(7\) terms:
\[S_7 = \frac{7}{2}[2(18) + 6(18)] = \frac{7}{2}(144) = 504\]
Exercise 5.3 — Q32
32. How many terms of the AP: \(-15, -13, -11, \dots\) are needed to make the sum \(-55\)? Explain the reason for double answer.
Solution
\[-55 = \frac{n}{2}[2(-15) + (n – 1)2] \implies -110 = n(2n – 32) \implies n^2 – 16n + 55 = 0\]
\[(n – 5)(n – 11) = 0 \implies n = 5 \text{ or } n = 11\]
Reason: Terms from the \(6^{\text{th}}\) to the \(11^{\text{th}}\) are \(-5, -3, -1, 1, 3, 5\), whose sum is \(0\). Thus, adding them leaves the sum unchanged.
Exercise 5.3 — Q33
33. The sum of first \(n\) terms of an AP (\(a = 8, d = 20\)) equals the sum of first \(2n\) terms of another AP (\(a = -30, d = 8\)). Find \(n\).
Solution
\[\frac{n}{2}[16 + (n – 1)20] = \frac{2n}{2}[-60 + (2n – 1)8]\]
\[\frac{n}{2}(20n – 4) = n(16n – 68)\]
\[10n – 2 = 16n – 68 \implies 6n = 66 \implies n = 11\]
Exercise 5.3 — Q34
34. Kanika puts \(\text{Re } 1\) on Day 1, \(\text{Rs } 2\) on Day 2, up to Jan 31st into her bank. She spent \(\text{Rs } 204\) and has \(\text{Rs } 100\) left. Find her pocket money.
Solution
\[S_{31} = \frac{31 \times 32}{2} = 496\]
\[\text{Total} = 496 + 204 + 100 = \text{Rs } 800\]
Exercise 5.3 — Q35
35. Yasmeen saves \(\text{Rs } 32\) in month 1, \(\text{Rs } 36\) in month 2, and \(\text{Rs } 40\) in month 3. In how many months will she save \(\text{Rs } 2000\)?
Solution
\[\frac{n}{2}[64 + (n – 1)4] = 2000 \implies n(2n + 30) = 2000 \implies n^2 + 15n – 1000 = 0\]
\[(n – 25)(n + 40) = 0 \implies n = 25 \text{ months}\]
Exercise 5.4: Long Answer Questions
Exercise 5.4 — Q1
1. The sum of the first \(5\) terms and first \(7\) terms of an AP is \(167\). If the sum of first \(10\) terms is \(235\), find the sum of its first \(20\) terms.
Solution
\[S_5 + S_7 = \frac{5}{2}(2a + 4d) + \frac{7}{2}(2a + 6d) = (5a + 10d) + (7a + 21d) = 12a + 31d = 167\]
\[S_{10} = 5(2a + 9d) = 235 \implies 2a + 9d = 47 \implies 12a + 54d = 282\]
Subtracting gives \(23d = 115 \implies d = 5\), which leads to \(2a = 47 – 45 = 2 \implies a = 1\).
\[S_{20} = \frac{20}{2}[2(1) + 19(5)] = 10(2 + 95) = 970\]
Exercise 5.4 — Q2
2. Find the sum of integers:
(i) between \(1\) and \(500\) which are multiples of \(2\) and \(5\)
(ii) from \(1\) to \(500\) which are multiples of \(2\) and \(5\)
(iii) from \(1\) to \(500\) which are multiples of \(2\) or \(5\)
Solution
(i) Between 1 and 500 (\(\text{LCM} = 10\)): \(10, 20, \dots, 490\) (\(n = 49\)):
\[S = \frac{49}{2}(10 + 490) = 49(250) = 12250\]
(ii) From 1 to 500: \(10, 20, \dots, 500\) (\(n = 50\)):
\[S = \frac{50}{2}(10 + 500) = 25(510) = 12750\]
(iii) Multiples of 2 or 5:
• Multiples of 2: \(2, 4, \dots, 500 \implies S_2 = \frac{250}{2}(2 + 500) = 62750\)
• Multiples of 5: \(5, 10, \dots, 500 \implies S_5 = \frac{100}{2}(5 + 500) = 25250\)
• Common (multiples of 10): \(S_{10} = 12750\)
\[S = S_2 + S_5 – S_{10} = 62750 + 25250 – 12750 = 75250\]
Exercise 5.4 — Q3
3. The eighth term of an AP is half its second term and eleventh term exceeds one-third of its fourth term by \(1\). Find the \(15^{\text{th}}\) term.
Solution
\[a + 7d = \frac{1}{2}(a + d) \implies 2a + 14d = a + d \implies a = -13d\]
\[(a + 10d) – \frac{1}{3}(a + 3d) = 1 \implies 3a + 30d – a – 3d = 3 \implies 2a + 27d = 3\]
Substituting \(a = -13d\):
\[2(-13d) + 27d = 3 \implies d = 3 \implies a = -39\]
\[a_{15} = -39 + 14(3) = 3\]
Exercise 5.4 — Q4
4. An AP consists of \(37\) terms. The sum of the three middle-most terms is \(225\) and sum of the last three is \(429\). Find the AP.
Solution
Middle terms are \(18^{\text{th}}, 19^{\text{th}}, 20^{\text{th}}\):
\[(a + 17d) + (a + 18d) + (a + 19d) = 3a + 54d = 225 \implies a + 18d = 75\]
Last three terms are \(35^{\text{th}}, 36^{\text{th}}, 37^{\text{th}}\):
\[(a + 34d) + (a + 35d) + (a + 36d) = 3a + 105d = 429 \implies a + 35d = 143\]
Subtracting yields \(17d = 68 \implies d = 4\) and \(a = 75 – 72 = 3\).
AP: \(\mathbf{3, 7, 11, 15, \dots, 147}\)
Exercise 5.4 — Q5
5. Find the sum of integers between \(100\) and \(200\) that are: (i) divisible by \(9\), (ii) not divisible by \(9\).
Solution
(i) Divisible by 9: \(108, 117, \dots, 198\) (\(n = 11\)):
\[S = \frac{11}{2}(108 + 198) = \frac{11}{2}(306) = 1683\]
(ii) Not divisible by 9: All integers between 100 and 200 are \(101, 102, \dots, 199\) (\(99\) terms):
\[S_{\text{total}} = \frac{99}{2}(101 + 199) = \frac{99}{2}(300) = 14850\]
\[S_{\text{not div}} = 14850 – 1683 = 13167\]
Exercise 5.4 — Q6
6. The ratio of the \(11^{\text{th}}\) term to the \(18^{\text{th}}\) term of an AP is \(2 : 3\). Find the ratio of the \(5^{\text{th}}\) to the \(21^{\text{st}}\) term, and ratio of the sum of the first \(5\) terms to the sum of the first \(21\) terms.
Solution
\[\frac{a + 10d}{a + 17d} = \frac{2}{3} \implies 3a + 30d = 2a + 34d \implies a = 4d\]
\[\frac{a_5}{a_{21}} = \frac{a + 4d}{a + 20d} = \frac{4d + 4d}{4d + 20d} = \frac{8d}{24d} = \mathbf{\frac{1}{3}}\]
\[\frac{S_5}{S_{21}} = \frac{\frac{5}{2}[2(4d) + 4d]}{\frac{21}{2}[2(4d) + 20d]} = \frac{5(12d)}{21(28d)} = \frac{60}{588} = \mathbf{\frac{5}{49}}\]
Exercise 5.4 — Q7
7. Show that the sum of an AP whose first term is \(a\), second term is \(b\), and last term is \(c\), is equal to \(\frac{(a + c)(b + c – 2a)}{2(b – a)}\).
Proof
Common difference \(d = b – a\). Last term \(c = a + (n – 1)(b – a)\):
\[n – 1 = \frac{c – a}{b – a} \implies n = 1 + \frac{c – a}{b – a} = \frac{b – a + c – a}{b – a} = \frac{b + c – 2a}{b – a}\]
Using the sum formula \(S_n = \frac{n}{2}(a + l)\) where \(l = c\):
\[S_n = \frac{1}{2} \cdot \frac{b + c – 2a}{b – a} \cdot (a + c) = \frac{(a + c)(b + c – 2a)}{2(b – a)}\]
Exercise 5.4 — Q8
8. Solve the equation: \(-4 + (-1) + 2 + \dots + x = 437\).
Solution
\(a = -4, d = 3\), \(x = -4 + (n – 1)3 = 3n – 7\):
\[S_n = \frac{n}{2}[-4 + 3n – 7] = 437 \implies n(3n – 11) = 874\]
\[3n^2 – 11n – 874 = 0\]
\[n = \frac{11 \pm \sqrt{121 – 4(3)(-874)}}{6} = \frac{11 \pm \sqrt{10609}}{6} = \frac{11 \pm 103}{6} \implies n = 19\]
\[x = 3(19) – 7 = 50\]
Exercise 5.4 — Q9
9. Jaspal Singh repays his loan of \(\text{Rs } 118000\) starting with \(\text{Rs } 1000\), increasing by \(\text{Rs } 100\) each month. What is paid in the \(30^{\text{th}}\) instalment and what remains after the \(30^{\text{th}}\) instalment?
Solution
\[a_{30} = 1000 + 29(100) = \mathbf{\text{Rs } 3900}\]
\[S_{30} = 15[2000 + 2900] = 15(4900) = \text{Rs } 73500\]
\[\text{Balance Left} = 118000 – 73500 = \mathbf{\text{Rs } 44500}\]
Exercise 5.4 — Q10
10. 27 flags are placed at intervals of \(2\text{ m}\) stored at the middle flag. Ruchi carries only one flag at a time, starting from storage and returning to collect books. Find total distance covered and maximum distance carrying a flag.
Solution
Storage is at the \(14^{\text{th}}\) flag (middle). 13 flags to the left and 13 flags to the right.
For one side (distances round trip: \(4, 8, 12, \dots\) to 13 terms):
\[S = \frac{13}{2}[8 + 12(4)] = \frac{13}{2}(56) = 364\text{ m}\]
\[\text{Total distance (both sides)} = 2 \times 364 = \mathbf{728\text{ m}}\]
\[\text{Max distance carrying a flag} = 13 \times 2 = \mathbf{26\text{ m}}\]
