Chapter 5: Arithmetic Progressions
Exercise 5.1
Q1
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.
Solution:
(i) Let \(a_n\) be the fare for the \(n^{\text{th}}\) km.
Fare for 1st km \(= 15\)
Fare for 2nd km \(= 15 + 8 = 23\)
Fare for 3rd km \(= 23 + 8 = 31\)
List of numbers: 15, 23, 31, ...
Since the difference between consecutive terms is constant (\(d = 8\)), it is an AP.
(ii) Let the initial volume of air be \(V\).
After 1st removal, air remaining \(= V - \frac{1}{4}V = \frac{3}{4}V\).
After 2nd removal, air remaining \(= \frac{3}{4}V - \frac{1}{4}\left(\frac{3}{4}V\right) = \frac{3}{4}V \left(1 - \frac{1}{4}\right) = \left(\frac{3}{4}\right)^2 V\).
List: \(V, \frac{3}{4}V, \left(\frac{3}{4}\right)^2 V, \dots\)
The difference between terms is not constant. It is not an AP.
(iii) Cost for 1st metre \(= 150\).
Cost for 2nd metre \(= 150 + 50 = 200\).
Cost for 3rd metre \(= 200 + 50 = 250\).
List: 150, 200, 250, ...
Difference is constant (\(50\)). It is an AP.
(iv) Amount after 1st year \(= 10000\left(1 + \frac{8}{100}\right)\).
Amount after 2nd year \(= 10000\left(1 + \frac{8}{100}\right)^2\).
This forms a Geometric Progression, not Arithmetic. It is not an AP.
Q2
Write first four terms of the AP, when the first term \(a\) and the common difference \(d\) are given as follows:
(i) \(a = 10, d = 10\)
(ii) \(a = -2, d = 0\)
(iii) \(a = 4, d = -3\)
(iv) \(a = -1, d = \frac{1}{2}\)
(v) \(a = -1.25, d = -0.25\)
(i) \(a = 10, d = 10\)
(ii) \(a = -2, d = 0\)
(iii) \(a = 4, d = -3\)
(iv) \(a = -1, d = \frac{1}{2}\)
(v) \(a = -1.25, d = -0.25\)
Solution:
(i) \(a=10, d=10\)
Terms: \(10, 10+10, 20+10, 30+10 \Rightarrow 10, 20, 30, 40\).
(ii) \(a=-2, d=0\)
Terms: \(-2, -2, -2, -2\).
(iii) \(a=4, d=-3\)
Terms: \(4, 4-3, 1-3, -2-3 \Rightarrow 4, 1, -2, -5\).
(iv) \(a=-1, d=0.5\)
Terms: \(-1, -0.5, 0, 0.5\).
(v) \(a=-1.25, d=-0.25\)
Terms: \(-1.25, -1.50, -1.75, -2.00\).
Q3
For the following APs, write the first term and the common difference:
(i) \(3, 1, -1, -3, \dots\)
(ii) \(-5, -1, 3, 7, \dots\)
(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots\)
(iv) \(0.6, 1.7, 2.8, 3.9, \dots\)
(i) \(3, 1, -1, -3, \dots\)
(ii) \(-5, -1, 3, 7, \dots\)
(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots\)
(iv) \(0.6, 1.7, 2.8, 3.9, \dots\)
Solution:
(i) First term \(a = 3\).
Common difference \(d = 1 - 3 = -2\).
(ii) First term \(a = -5\).
Common difference \(d = -1 - (-5) = -1 + 5 = 4\).
(iii) First term \(a = \frac{1}{3}\).
Common difference \(d = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}\).
(iv) First term \(a = 0.6\).
Common difference \(d = 1.7 - 0.6 = 1.1\).
Q4
Which of the following are APs? If they form an AP, find the common difference \(d\) and write three more terms.
(i) \(2, 4, 8, 16, \dots\) (ii) \(2, \frac{5}{2}, 3, \frac{7}{2}, \dots\) (iii) \(-1.2, -3.2, -5.2, -7.2, \dots\)
(iv) \(-10, -6, -2, 2, \dots\) (v) \(3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots\)
(vi) \(0.2, 0.22, 0.222, 0.2222, \dots\) (vii) \(0, -4, -8, -12, \dots\)
(viii) \(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots\) (ix) \(1, 3, 9, 27, \dots\)
(x) \(a, 2a, 3a, 4a, \dots\) (xi) \(a, a^2, a^3, a^4, \dots\)
(xii) \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots\) (xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots\)
(xiv) \(1^2, 3^2, 5^2, 7^2, \dots\) (xv) \(1^2, 5^2, 7^2, 73, \dots\)
(i) \(2, 4, 8, 16, \dots\) (ii) \(2, \frac{5}{2}, 3, \frac{7}{2}, \dots\) (iii) \(-1.2, -3.2, -5.2, -7.2, \dots\)
(iv) \(-10, -6, -2, 2, \dots\) (v) \(3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots\)
(vi) \(0.2, 0.22, 0.222, 0.2222, \dots\) (vii) \(0, -4, -8, -12, \dots\)
(viii) \(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots\) (ix) \(1, 3, 9, 27, \dots\)
(x) \(a, 2a, 3a, 4a, \dots\) (xi) \(a, a^2, a^3, a^4, \dots\)
(xii) \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots\) (xiii) \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots\)
(xiv) \(1^2, 3^2, 5^2, 7^2, \dots\) (xv) \(1^2, 5^2, 7^2, 73, \dots\)
Solution:
(i) \(4-2=2, 8-4=4\). Not an AP.
(ii) \(\frac{5}{2} - 2 = 0.5\), \(3 - \frac{5}{2} = 0.5\). Yes, AP. \(d=0.5\). Next terms: \(4, 4.5, 5\).
(iii) \(-3.2 - (-1.2) = -2\), \(-5.2 - (-3.2) = -2\). Yes, AP. \(d=-2\). Next terms: \(-9.2, -11.2, -13.2\).
(iv) \(-6 - (-10) = 4\), \(-2 - (-6) = 4\). Yes, AP. \(d=4\). Next terms: \(6, 10, 14\).
(v) Difference is \(\sqrt{2}\). Yes, AP. Next terms: \(3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}\).
(vi) \(0.22 - 0.2 = 0.02\), \(0.222 - 0.22 = 0.002\). Not an AP.
(vii) \(-4-0=-4\). Yes, AP. \(d=-4\). Next terms: \(-16, -20, -24\).
(viii) Difference is 0. Yes, AP. \(d=0\). Next terms: \(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}\).
(ix) \(3-1=2, 9-3=6\). Not an AP.
(x) \(2a-a=a, 3a-2a=a\). Yes, AP. \(d=a\). Next terms: \(5a, 6a, 7a\).
(xi) \(a^2-a \neq a^3-a^2\). Not an AP.
(xii) Sequence: \(\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}\). \(d=\sqrt{2}\). Yes, AP. Next terms: \(5\sqrt{2} (\sqrt{50}), 6\sqrt{2} (\sqrt{72}), 7\sqrt{2} (\sqrt{98})\).
(xiii) \(\sqrt{6}-\sqrt{3} \neq \sqrt{9}-\sqrt{6}\). Not an AP.
(xiv) \(9-1=8, 25-9=16\). Not an AP.
(xv) Sequence: 1, 25, 49, 73... Diff: \(25-1=24, 49-25=24\). Yes, AP. \(d=24\). Next terms: \(97, 121, 145\).
Exercise 5.2
Q1
Fill in the blanks in the following table, given that \(a\) is the first term, \(d\) the common difference and \(a_n\) the nth term of the AP:
(i) \(a=7, d=3, n=8, a_n=\dots\)
(ii) \(a=-18, d=\dots, n=10, a_n=0\)
(iii) \(a=\dots, d=-3, n=18, a_n=-5\)
(iv) \(a=-18.9, d=2.5, n=\dots, a_n=3.6\)
(v) \(a=3.5, d=0, n=105, a_n=\dots\)
(i) \(a=7, d=3, n=8, a_n=\dots\)
(ii) \(a=-18, d=\dots, n=10, a_n=0\)
(iii) \(a=\dots, d=-3, n=18, a_n=-5\)
(iv) \(a=-18.9, d=2.5, n=\dots, a_n=3.6\)
(v) \(a=3.5, d=0, n=105, a_n=\dots\)
Solution: Formula: \(a_n = a + (n-1)d\)
(i) \(a_n = 7 + (8-1)3 = 7 + 21 = 28\).
(ii) \(0 = -18 + (10-1)d \Rightarrow 18 = 9d \Rightarrow d = 2\).
(iii) \(-5 = a + (18-1)(-3) \Rightarrow -5 = a - 51 \Rightarrow a = 46\).
(iv) \(3.6 = -18.9 + (n-1)2.5 \Rightarrow 22.5 = (n-1)2.5 \Rightarrow n-1 = 9 \Rightarrow n = 10\).
(v) \(a_n = 3.5 + (105-1)(0) = 3.5\).
Q2
Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, ..., is
(ii) 11th term of the AP: -3, -1/2, 2, ..., is
(i) 30th term of the AP: 10, 7, 4, ..., is
(ii) 11th term of the AP: -3, -1/2, 2, ..., is
Solution:
(i) \(a=10, d=7-10=-3, n=30\).
\[ a_{30} = 10 + (29)(-3) = 10 - 87 = -77 \]
Correct choice: (C)
(ii) \(a=-3, d=-\frac{1}{2} - (-3) = 2.5, n=11\).
\[ a_{11} = -3 + (10)(2.5) = -3 + 25 = 22 \]
Correct choice: (B)
Q3
In the following APs, find the missing terms in the boxes:
(i) \( 2, \boxed{?}, 26 \)
(ii) \( \boxed{?}, 13, \boxed{?}, 3 \)
(iii) \( 5, \boxed{?}, \boxed{?}, 9\frac{1}{2} \)
(iv) \( -4, \boxed{?}, \boxed{?}, \boxed{?}, \boxed{?}, 6 \)
(v) \( \boxed{?}, 38, \boxed{?}, \boxed{?}, \boxed{?}, -22 \)
(i) \( 2, \boxed{?}, 26 \)
(ii) \( \boxed{?}, 13, \boxed{?}, 3 \)
(iii) \( 5, \boxed{?}, \boxed{?}, 9\frac{1}{2} \)
(iv) \( -4, \boxed{?}, \boxed{?}, \boxed{?}, \boxed{?}, 6 \)
(v) \( \boxed{?}, 38, \boxed{?}, \boxed{?}, \boxed{?}, -22 \)
Solution:
(i) \( 2, \boxed{?}, 26 \)
We know that for an AP, the middle term is the arithmetic mean of the adjacent terms.
\[ \text{Missing term} = \frac{2 + 26}{2} = \frac{28}{2} = 14 \]
Answer: 14
(ii) \( \boxed{?}, 13, \boxed{?}, 3 \)
Let the terms be \(a, a+d, a+2d, a+3d\).
\(a_2 = a + d = 13\) ... (1)
\(a_4 = a + 3d = 3\) ... (2)
Subtracting (1) from (2): \(2d = -10 \Rightarrow d = -5\).
Substitute \(d = -5\) in (1): \(a - 5 = 13 \Rightarrow a = 18\).
Third term \(a_3 = a + 2d = 18 + 2(-5) = 8\).
Answer: 18, 8
(iii) \( 5, \boxed{?}, \boxed{?}, 9\frac{1}{2} \)
\(a = 5\), \(a_4 = 9.5\).
\(a + 3d = 9.5 \Rightarrow 5 + 3d = 9.5 \Rightarrow 3d = 4.5 \Rightarrow d = 1.5\).
\(a_2 = 5 + 1.5 = 6.5\)
\(a_3 = 6.5 + 1.5 = 8\)
Answer: 6.5, 8
(iv) \( -4, \boxed{?}, \boxed{?}, \boxed{?}, \boxed{?}, 6 \)
\(a = -4\), \(a_6 = 6\).
\(a + 5d = 6 \Rightarrow -4 + 5d = 6 \Rightarrow 5d = 10 \Rightarrow d = 2\).
Terms: \(-4+2=-2\), \(-2+2=0\), \(0+2=2\), \(2+2=4\).
Answer: -2, 0, 2, 4
(v) \( \boxed{?}, 38, \boxed{?}, \boxed{?}, \boxed{?}, -22 \)
\(a_2 = 38 \Rightarrow a + d = 38\) ... (1)
\(a_6 = -22 \Rightarrow a + 5d = -22\) ... (2)
Subtract (1) from (2): \(4d = -60 \Rightarrow d = -15\).
Substitute \(d = -15\) in (1): \(a - 15 = 38 \Rightarrow a = 53\).
Terms: \(53, 38, 23, 8, -7, -22\).
Answer: 53, 23, 8, -7
Q4
Which term of the AP: 3, 8, 13, 18, ..., is 78?
Solution:
\(a=3, d=5, a_n=78\).
\[ 78 = 3 + (n-1)5 \]
\[ 75 = 5(n-1) \Rightarrow 15 = n-1 \Rightarrow n = 16 \]
Answer: 16th term.
Q5
Find the number of terms in each of the following APs:
(i) 7, 13, 19, ..., 205
(ii) 18, \(15\frac{1}{2}\), 13, ..., -47
(i) 7, 13, 19, ..., 205
(ii) 18, \(15\frac{1}{2}\), 13, ..., -47
Solution:
(i) \(a=7, d=6, a_n=205\).
\[ 205 = 7 + (n-1)6 \Rightarrow 198 = 6(n-1) \Rightarrow 33 = n-1 \Rightarrow n = 34 \]
(ii) \(a=18, d=15.5-18=-2.5, a_n=-47\).
\[ -47 = 18 + (n-1)(-2.5) \]
\[ -65 = (n-1)(-2.5) \Rightarrow n-1 = 26 \Rightarrow n = 27 \]
Q6
Check whether -150 is a term of the AP: 11, 8, 5, 2...
Solution:
\(a=11, d=-3\). Assume \(a_n = -150\).
\[ -150 = 11 + (n-1)(-3) \]
\[ -161 = (n-1)(-3) \Rightarrow n-1 = \frac{161}{3} \]
\( \frac{161}{3} \) is not an integer. Therefore, -150 is not a term of the AP.
Q7
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
\(a_{11} = a + 10d = 38\)
\(a_{16} = a + 15d = 73\)
Subtracting: \(5d = 35 \Rightarrow d = 7\).
Substitute \(d=7\): \(a + 70 = 38 \Rightarrow a = -32\).
\[ a_{31} = a + 30d = -32 + 30(7) = -32 + 210 = 178 \]
Q8
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
\(a_3 = a + 2d = 12\)
\(a_{50} = a + 49d = 106\)
Subtracting: \(47d = 94 \Rightarrow d = 2\).
Substitute \(d=2\): \(a + 4 = 12 \Rightarrow a = 8\).
\[ a_{29} = a + 28d = 8 + 28(2) = 8 + 56 = 64 \]
Q9
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
\(a_3 = a + 2d = 4\)
\(a_9 = a + 8d = -8\)
Subtracting: \(6d = -12 \Rightarrow d = -2\).
Substitute \(d=-2\): \(a - 4 = 4 \Rightarrow a = 8\).
Let \(a_n = 0\):
\[ 8 + (n-1)(-2) = 0 \Rightarrow 2(n-1) = 8 \Rightarrow n-1 = 4 \Rightarrow n = 5 \]
Answer: 5th term.
Q10
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Given: \(a_{17} - a_{10} = 7\)
\[ (a + 16d) - (a + 9d) = 7 \]
\[ 7d = 7 \Rightarrow d = 1 \]
Answer: 1.
Q11
Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?
Solution:
\(a=3, d=12\).
\(a_{54} = 3 + 53(12) = 3 + 636 = 639\).
Required term value \(= 639 + 132 = 771\).
\[ 771 = 3 + (n-1)12 \Rightarrow 768 = 12(n-1) \Rightarrow 64 = n-1 \Rightarrow n = 65 \]
Answer: 65th term.
Q12
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let the first terms be \(a\) and \(A\), common difference \(d\).
Diff of 100th terms: \((a + 99d) - (A + 99d) = a - A = 100\).
Diff of 1000th terms: \((a + 999d) - (A + 999d) = a - A\).
Since \(a - A = 100\), the difference is 100.
Q13
How many three-digit numbers are divisible by 7?
Solution:
AP: 105, 112, ..., 994. \(a=105, d=7, a_n=994\).
\[ 994 = 105 + (n-1)7 \]
\[ 889 = 7(n-1) \Rightarrow 127 = n-1 \Rightarrow n = 128 \]
Q14
How many multiples of 4 lie between 10 and 250?
Solution:
AP: 12, 16, ..., 248. \(a=12, d=4, a_n=248\).
\[ 248 = 12 + (n-1)4 \]
\[ 236 = 4(n-1) \Rightarrow 59 = n-1 \Rightarrow n = 60 \]
Q15
For what value of \(n\), are the \(n\)th terms of two APs: 63, 65, 67, ... and 3, 10, 17, ... equal?
Solution:
AP1: \(63 + (n-1)2\).
AP2: \(3 + (n-1)7\).
\[ 63 + 2n - 2 = 3 + 7n - 7 \]
\[ 61 + 2n = 7n - 4 \Rightarrow 65 = 5n \Rightarrow n = 13 \]
Q16
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
\(a_7 - a_5 = 12 \Rightarrow (a+6d) - (a+4d) = 12 \Rightarrow 2d = 12 \Rightarrow d = 6\).
\(a_3 = 16 \Rightarrow a + 2(6) = 16 \Rightarrow a + 12 = 16 \Rightarrow a = 4\).
AP: 4, 10, 16, 22, ...
Q17
Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.
Solution:
Reverse the AP: 253, 248, ..., 3.
\(a=253, d=-5\).
\[ a_{20} = 253 + 19(-5) = 253 - 95 = 158 \]
Q18
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 of the AP.
Solution:
1) \(a_4 + a_8 = 24 \Rightarrow (a+3d) + (a+7d) = 24 \Rightarrow 2a + 10d = 24 \Rightarrow a + 5d = 12\).
2) \(a_6 + a_{10} = 44 \Rightarrow (a+5d) + (a+9d) = 44 \Rightarrow 2a + 14d = 44 \Rightarrow a + 7d = 22\).
Subtracting eq(1) from eq(2): \(2d = 10 \Rightarrow d = 5\).
\(a + 25 = 12 \Rightarrow a = -13\).
First three terms: -13, -8, -3.
Q19
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Solution:
\(a=5000, d=200, a_n=7000\).
\[ 7000 = 5000 + (n-1)200 \]
\[ 2000 = 200(n-1) \Rightarrow 10 = n-1 \Rightarrow n = 11 \]
In the 11th year (i.e., 2005).
Q20
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find \(n\).
Solution:
\(a=5, d=1.75, a_n=20.75\).
\[ 20.75 = 5 + (n-1)1.75 \]
\[ 15.75 = 1.75(n-1) \Rightarrow n-1 = \frac{1575}{175} = 9 \Rightarrow n = 10 \]
Exercise 5.3
Q1
Find the sum of the following APs:
(i) 2, 7, 12, ..., to 10 terms.
(ii) -37, -33, -29, ..., to 12 terms.
(iii) 0.6, 1.7, 2.8, ..., to 100 terms.
(iv) \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots\), to 11 terms.
(i) 2, 7, 12, ..., to 10 terms.
(ii) -37, -33, -29, ..., to 12 terms.
(iii) 0.6, 1.7, 2.8, ..., to 100 terms.
(iv) \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots\), to 11 terms.
Solution: Formula: \(S_n = \frac{n}{2}[2a + (n-1)d]\)
(i) \(a=2, d=5, n=10\).
\(S_{10} = \frac{10}{2}[2(2) + 9(5)] = 5[4 + 45] = 5(49) = 245\).
(ii) \(a=-37, d=4, n=12\).
\(S_{12} = \frac{12}{2}[2(-37) + 11(4)] = 6[-74 + 44] = 6(-30) = -180\).
(iii) \(a=0.6, d=1.1, n=100\).
\(S_{100} = \frac{100}{2}[2(0.6) + 99(1.1)] = 50[1.2 + 108.9] = 50(110.1) = 5505\).
(iv) \(a=\frac{1}{15}, d=\frac{1}{12}-\frac{1}{15} = \frac{1}{60}, n=11\).
\(S_{11} = \frac{11}{2}[2(\frac{1}{15}) + 10(\frac{1}{60})] = \frac{11}{2}[\frac{2}{15} + \frac{1}{6}] = \frac{11}{2}[\frac{4+5}{30}] = \frac{11}{2}(\frac{9}{30}) = \frac{33}{20}\).
Q2
Find the sums given below:
(i) \(7 + 10\frac{1}{2} + 14 + \dots + 84\)
(ii) \(34 + 32 + 30 + \dots + 10\)
(iii) \(-5 + (-8) + (-11) + \dots + (-230)\)
(i) \(7 + 10\frac{1}{2} + 14 + \dots + 84\)
(ii) \(34 + 32 + 30 + \dots + 10\)
(iii) \(-5 + (-8) + (-11) + \dots + (-230)\)
Solution: Formula: \(S_n = \frac{n}{2}(a + l)\)
(i) \(a=7, d=3.5, l=84\). Find \(n\):
\(84 = 7 + (n-1)3.5 \Rightarrow 77 = (n-1)3.5 \Rightarrow n-1 = 22 \Rightarrow n = 23\).
\(S_{23} = \frac{23}{2}(7 + 84) = \frac{23}{2}(91) = 1046.5\).
(ii) \(a=34, d=-2, l=10\).
\(10 = 34 + (n-1)(-2) \Rightarrow -24 = -2(n-1) \Rightarrow n-1 = 12 \Rightarrow n = 13\).
\(S_{13} = \frac{13}{2}(34 + 10) = \frac{13}{2}(44) = 13 \times 22 = 286\).
(iii) \(a=-5, d=-3, l=-230\).
\(-230 = -5 + (n-1)(-3) \Rightarrow -225 = -3(n-1) \Rightarrow n-1 = 75 \Rightarrow n = 76\).
\(S_{76} = \frac{76}{2}(-5 - 230) = 38(-235) = -8930\).
Q3
In an AP:
(i) given \(a=5, d=3, a_n=50\), find \(n\) and \(S_n\).
(ii) given \(a=7, a_{13}=35\), find \(d\) and \(S_{13}\).
(iii) given \(a_{12}=37, d=3\), find \(a\) and \(S_{12}\).
(iv) given \(a_3=15, S_{10}=125\), find \(d\) and \(a_{10}\).
(v) given \(d=5, S_9=75\), find \(a\) and \(a_9\).
(vi) given \(a=2, d=8, S_n=90\), find \(n\) and \(a_n\).
(vii) given \(a=8, a_n=62, S_n=210\), find \(n\) and \(d\).
(viii) given \(a_n=4, d=2, S_n=-14\), find \(n\) and \(a\).
(ix) given \(a=3, n=8, S=192\), find \(d\).
(x) given \(l=28, S=144\), and there are total 9 terms. Find \(a\).
(i) given \(a=5, d=3, a_n=50\), find \(n\) and \(S_n\).
(ii) given \(a=7, a_{13}=35\), find \(d\) and \(S_{13}\).
(iii) given \(a_{12}=37, d=3\), find \(a\) and \(S_{12}\).
(iv) given \(a_3=15, S_{10}=125\), find \(d\) and \(a_{10}\).
(v) given \(d=5, S_9=75\), find \(a\) and \(a_9\).
(vi) given \(a=2, d=8, S_n=90\), find \(n\) and \(a_n\).
(vii) given \(a=8, a_n=62, S_n=210\), find \(n\) and \(d\).
(viii) given \(a_n=4, d=2, S_n=-14\), find \(n\) and \(a\).
(ix) given \(a=3, n=8, S=192\), find \(d\).
(x) given \(l=28, S=144\), and there are total 9 terms. Find \(a\).
Solution:
(i) \(50 = 5 + (n-1)3 \Rightarrow 45 = 3(n-1) \Rightarrow n=16\).
\(S_{16} = \frac{16}{2}(5+50) = 8(55) = 440\).
(ii) \(35 = 7 + 12d \Rightarrow 28 = 12d \Rightarrow d = \frac{7}{3}\).
\(S_{13} = \frac{13}{2}(7+35) = \frac{13}{2}(42) = 13 \times 21 = 273\).
(iii) \(37 = a + 11(3) \Rightarrow a = 37 - 33 = 4\).
\(S_{12} = \frac{12}{2}(4+37) = 6(41) = 246\).
(iv) \(a+2d=15\) ... (1). \(S_{10} = \frac{10}{2}[2a+9d] = 125 \Rightarrow 2a+9d=25\) ... (2).
Solving (1) and (2): \(d=-1, a=17\). \(a_{10} = 17 + 9(-1) = 8\).
(v) \(S_9 = \frac{9}{2}[2a + 8(5)] = 75 \Rightarrow 9(a+20) = 75 \Rightarrow 3a+60=25 \Rightarrow a = -\frac{35}{3}\).
\(a_9 = -\frac{35}{3} + 8(5) = \frac{85}{3}\).
(vi) \(90 = \frac{n}{2}[4 + (n-1)8] \Rightarrow 180 = n(8n-4) \Rightarrow 2n^2 - n - 45 = 0\).
Solving quadratic: \(n=5\) (since \(n\) is positive). \(a_5 = 2 + 4(8) = 34\).
(vii) \(210 = \frac{n}{2}(8+62) \Rightarrow 420 = 70n \Rightarrow n=6\).
\(62 = 8 + 5d \Rightarrow 54 = 5d \Rightarrow d = \frac{54}{5} = 10.8\).
(viii) \(a_n=4 \Rightarrow a+2n-2=4 \Rightarrow a=6-2n\).
\(-14 = \frac{n}{2}(a+4) \Rightarrow -28 = n(6-2n+4) \Rightarrow n^2-5n-14=0\).
\(n=7\) (positive). \(a = 6 - 2(7) = -8\).
(ix) \(192 = \frac{8}{2}[2(3) + 7d] \Rightarrow 192 = 4(6+7d) \Rightarrow 48 = 6+7d \Rightarrow d=6\).
(x) \(144 = \frac{9}{2}(a+28) \Rightarrow 288 = 9(a+28) \Rightarrow 32 = a+28 \Rightarrow a=4\).
Q4
How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
Solution:
\(a=9, d=8, S_n=636\).
\(636 = \frac{n}{2}[2(9) + (n-1)8] = \frac{n}{2}[18 + 8n - 8] = \frac{n}{2}[10 + 8n] = n(5 + 4n)\).
\(4n^2 + 5n - 636 = 0\).
Solving quadratic equation: \(n = \frac{-5 \pm \sqrt{25 - 4(4)(-636)}}{8} = \frac{-5 \pm \sqrt{10201}}{8} = \frac{-5 \pm 101}{8}\).
Taking positive root: \(n = \frac{96}{8} = 12\).
Answer: 12 terms.
Q5
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
\(a=5, l=45, S=400\).
\(S = \frac{n}{2}(a+l) \Rightarrow 400 = \frac{n}{2}(5+45) \Rightarrow 400 = 25n \Rightarrow n = 16\).
\(l = a + (n-1)d \Rightarrow 45 = 5 + 15d \Rightarrow 40 = 15d \Rightarrow d = \frac{40}{15} = \frac{8}{3}\).
Answer: \(n = 16, d = \frac{8}{3}\).
Q6
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
\(a=17, l=350, d=9\).
\(350 = 17 + (n-1)9 \Rightarrow 333 = 9(n-1) \Rightarrow 37 = n-1 \Rightarrow n = 38\).
\(S_{38} = \frac{38}{2}(17 + 350) = 19(367) = 6973\).
Answer: 38 terms, Sum = 6973.
Q7
Find the sum of first 22 terms of an AP in which \(d = 7\) and 22nd term is 149.
Solution:
\(a_{22} = a + 21d \Rightarrow 149 = a + 21(7) \Rightarrow 149 = a + 147 \Rightarrow a = 2\).
\(S_{22} = \frac{22}{2}(a + a_{22}) = 11(2 + 149) = 11(151) = 1661\).
Answer: 1661.
Q8
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
\(a_2 = 14, a_3 = 18\).
\(d = a_3 - a_2 = 18 - 14 = 4\).
\(a = a_2 - d = 14 - 4 = 10\).
\(S_{51} = \frac{51}{2}[2(10) + 50(4)] = \frac{51}{2}[20 + 200] = \frac{51}{2}(220) = 51 \times 110 = 5610\).
Answer: 5610.
Q9
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first \(n\) terms.
Solution:
\(S_7 = 49 \Rightarrow 7^2\). \(S_{17} = 289 \Rightarrow 17^2\).
The sum of the first \(n\) terms is \(n^2\).
Alternatively, solving equations:
\(\frac{7}{2}[2a+6d]=49 \Rightarrow a+3d=7\)
\(\frac{17}{2}[2a+16d]=289 \Rightarrow a+8d=17\)
Solving gives \(d=2, a=1\).
\(S_n = \frac{n}{2}[2(1) + (n-1)2] = \frac{n}{2}[2n] = n^2\).
Answer: \(n^2\).
Q10
Show that \(a_1, a_2, \dots, a_n, \dots\) form an AP where \(a_n\) is defined as below:
(i) \(a_n = 3 + 4n\)
(ii) \(a_n = 9 - 5n\)
Also find the sum of the first 15 terms in each case.
(i) \(a_n = 3 + 4n\)
(ii) \(a_n = 9 - 5n\)
Also find the sum of the first 15 terms in each case.
Solution:
(i) \(a_1 = 7, a_2 = 11, a_3 = 15\). Difference is 4. It is an AP with \(a=7, d=4\).
\(S_{15} = \frac{15}{2}[2(7) + 14(4)] = \frac{15}{2}[14 + 56] = \frac{15}{2}(70) = 525\).
(ii) \(a_1 = 4, a_2 = -1, a_3 = -6\). Difference is -5. It is an AP with \(a=4, d=-5\).
\(S_{15} = \frac{15}{2}[2(4) + 14(-5)] = \frac{15}{2}[8 - 70] = \frac{15}{2}(-62) = -465\).
Q11
If the sum of the first \(n\) terms of an AP is \(4n - n^2\), what is the first term (that is \(S_1\))? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the \(n\)th terms.
Solution:
\(S_n = 4n - n^2\).
\(S_1 = 4(1) - 1^2 = 3\). First term \(a_1 = 3\).
\(S_2 = 4(2) - 2^2 = 8 - 4 = 4\). (Sum of first two terms).
\(a_2 = S_2 - S_1 = 4 - 3 = 1\).
\(S_3 = 4(3) - 3^2 = 3\). \(a_3 = S_3 - S_2 = 3 - 4 = -1\).
AP: 3, 1, -1... Common difference \(d = -2\).
\(a_{10} = S_{10} - S_9 = (40 - 100) - (36 - 81) = -60 - (-45) = -15\).
\(a_n = S_n - S_{n-1} = (4n - n^2) - [4(n-1) - (n-1)^2] = 5 - 2n\).
Q12
Find the sum of the first 40 positive integers divisible by 6.
Solution:
AP: 6, 12, 18, ...
\(a=6, d=6, n=40\).
\(S_{40} = \frac{40}{2}[2(6) + 39(6)] = 20[12 + 234] = 20(246) = 4920\).
Q13
Find the sum of the first 15 multiples of 8.
Solution:
AP: 8, 16, 24, ...
\(a=8, d=8, n=15\).
\(S_{15} = \frac{15}{2}[2(8) + 14(8)] = \frac{15}{2}[16 + 112] = \frac{15}{2}(128) = 15 \times 64 = 960\).
Q14
Find the sum of the odd numbers between 0 and 50.
Solution:
AP: 1, 3, 5, ..., 49.
\(a=1, d=2, l=49\). Number of terms \(n = \frac{49-1}{2} + 1 = 25\).
\(S_{25} = \frac{25}{2}(1 + 49) = \frac{25}{2}(50) = 25 \times 25 = 625\).
Q15
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
AP: 200, 250, 300, ...
\(a=200, d=50, n=30\).
\(S_{30} = \frac{30}{2}[2(200) + 29(50)] = 15[400 + 1450] = 15(1850) = 27750\).
Answer: ₹ 27,750.
Q16
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let prizes be \(a, a-20, a-40, \dots\)
\(n=7, S_7=700, d=-20\).
\(700 = \frac{7}{2}[2a + 6(-20)] \Rightarrow 200 = 2a - 120 \Rightarrow 2a = 320 \Rightarrow a = 160\).
The prizes are: 160, 140, 120, 100, 80, 60, 40.
Q17
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying. There are three sections of each class. How many trees will be planted by the students?
Solution:
Trees by Class I = \(1 \times 3 = 3\).
Trees by Class II = \(2 \times 3 = 6\).
Trees by Class XII = \(12 \times 3 = 36\).
AP: 3, 6, ..., 36. \(n=12\).
\(S_{12} = \frac{12}{2}(3 + 36) = 6(39) = 234\).
Answer: 234 trees.
Q18
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take \(\pi = \frac{22}{7}\))
Solution:
Perimeter of semicircle \( = \pi r \).
Radii: 0.5, 1.0, 1.5, ...
Lengths: \(\pi(0.5), \pi(1.0), \pi(1.5), \dots\)
Total Length \(S_{13} = \pi [0.5 + 1.0 + \dots + \text{13 terms}]\).
The series inside is an AP with \(a=0.5, d=0.5, n=13\).
\(S_{AP} = \frac{13}{2}[2(0.5) + 12(0.5)] = \frac{13}{2}[1 + 6] = \frac{91}{2}\).
Total Length \( = \frac{22}{7} \times \frac{91}{2} = 11 \times 13 = 143\) cm.
Q19
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
Solution:
AP: 20, 19, 18... \(S_n = 200\). \(a=20, d=-1\).
\(200 = \frac{n}{2}[2(20) + (n-1)(-1)] \Rightarrow 400 = n(40 - n + 1) \Rightarrow 400 = 41n - n^2\).
\(n^2 - 41n + 400 = 0\).
\((n-16)(n-25) = 0 \Rightarrow n = 16 \text{ or } 25\).
If \(n=25, a_{25} = 20 + 24(-1) = -4\) (Not possible).
If \(n=16, a_{16} = 20 + 15(-1) = 5\).
Answer: 16 rows, 5 logs in the top row.
Q20
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
Solution:
Distance for 1st potato: \(2 \times 5 = 10\) m.
Distance for 2nd potato: \(2 \times (5+3) = 16\) m.
Distance for 3rd potato: \(2 \times (5+3+3) = 22\) m.
AP: 10, 16, 22... \(n=10\). \(a=10, d=6\).
\(S_{10} = \frac{10}{2}[2(10) + 9(6)] = 5[20 + 54] = 5(74) = 370\) m.
Answer: 370 m.
Exercise 5.4 (Optional)
Q1
Which term of the AP: 121, 117, 113, ..., is its first negative term? [Hint: Find \(n\) for \(a_n < 0\)]
Solution:
Given AP: 121, 117, 113, ...
First term \(a = 121\).
Common difference \(d = 117 - 121 = -4\).
We need to find the first negative term, i.e., \(a_n < 0\).
\[ a + (n-1)d < 0 \]
\[ 121 + (n-1)(-4) < 0 \]
\[ 121 - 4n + 4 < 0 \]
\[ 125 < 4n \]
\[ n > \frac{125}{4} \]
\[ n > 31.25 \]
The first integer greater than 31.25 is 32.
Answer: The 32nd term is the first negative term.
Q2
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Solution:
Let the AP be \(a, a+d, a+2d, \dots\)
Condition 1: Sum of 3rd and 7th terms is 6.
\[ a_3 + a_7 = 6 \]
\[ (a + 2d) + (a + 6d) = 6 \]
\[ 2a + 8d = 6 \Rightarrow a + 4d = 3 \Rightarrow a = 3 - 4d \quad \dots(1) \]
Condition 2: Product is 8.
\[ (a + 2d)(a + 6d) = 8 \]
Substitute \(a = 3 - 4d\) into this equation:
\[ (3 - 4d + 2d)(3 - 4d + 6d) = 8 \]
\[ (3 - 2d)(3 + 2d) = 8 \]
\[ 9 - 4d^2 = 8 \]
\[ 4d^2 = 1 \Rightarrow d^2 = \frac{1}{4} \Rightarrow d = \pm \frac{1}{2} \]
Case 1: When \(d = \frac{1}{2}\)
\[ a = 3 - 4(\frac{1}{2}) = 1 \]
\[ S_{16} = \frac{16}{2}[2(1) + 15(\frac{1}{2})] = 8[2 + 7.5] = 8(9.5) = 76 \]
Case 2: When \(d = -\frac{1}{2}\)
\[ a = 3 - 4(-\frac{1}{2}) = 5 \]
\[ S_{16} = \frac{16}{2}[2(5) + 15(-\frac{1}{2})] = 8[10 - 7.5] = 8(2.5) = 20 \]
Answer: The sum is either 76 or 20.
Q3
A ladder has rungs 25 cm apart (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are \(2\frac{1}{2}\) m apart, what is the length of the wood required for the rungs? [Hint: Number of rungs = \(\frac{250}{25} + 1\)]
Solution:
Total distance between top and bottom rungs = \(2\frac{1}{2}\) m = 2.5 m = 250 cm.
Distance between consecutive rungs = 25 cm.
\[ \text{Number of rungs } (n) = \frac{\text{Total distance}}{\text{Distance between rungs}} + 1 \]
\[ n = \frac{250}{25} + 1 = 10 + 1 = 11 \]
The lengths of the rungs form an AP because they decrease uniformly.
First term \(a = 45\) (bottom rung).
Last term \(l = 25\) (top rung).
Number of terms \(n = 11\).
Total length of wood = Sum of the AP
\[ S_{11} = \frac{n}{2}(a + l) \]
\[ S_{11} = \frac{11}{2}(45 + 25) = \frac{11}{2}(70) = 11 \times 35 = 385 \text{ cm} \]
Answer: The length of wood required is 385 cm.
Q4
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of \(x\) such that the sum of the numbers of the houses preceding the house numbered \(x\) is equal to the sum of the numbers of the houses following it. Find this value of \(x\). [Hint: \(S_{x-1} = S_{49} - S_x\)]
Solution:
The house numbers form an AP: 1, 2, 3, ..., 49.
Here \(a = 1\) and \(d = 1\).
We are given that sum of numbers before \(x\) equals sum of numbers after \(x\).
\[ S_{x-1} = S_{49} - S_x \]
Using the formula \(S_n = \frac{n}{2}(a + l)\) or \(\frac{n}{2}[2a + (n-1)d]\):
\[ \frac{x-1}{2}[1 + (x-1)] = \frac{49}{2}[1 + 49] - \frac{x}{2}[1 + x] \]
\[ \frac{x(x-1)}{2} = \frac{49(50)}{2} - \frac{x(x+1)}{2} \]
Multiply by 2 to clear fractions:
\[ x^2 - x = 2450 - (x^2 + x) \]
\[ x^2 - x = 2450 - x^2 - x \]
\[ 2x^2 = 2450 \]
\[ x^2 = 1225 \]
\[ x = \sqrt{1225} = 35 \]
Answer: The value of \(x\) is 35.
Q5
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\) m and a tread of \(\frac{1}{2}\) m (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint: Volume of concrete required to build the first step = \(\frac{1}{4} \times \frac{1}{2} \times 50\) \(m^3\)]
Solution:
We calculate the volume of concrete for each step.
Step 1:
Volume \(V_1 = \text{Length} \times \text{Breadth} \times \text{Height} = 50 \times \frac{1}{2} \times \frac{1}{4} = 6.25 \text{ m}^3\).
Step 2: The height increases by \(\frac{1}{4}\) m, but length and breadth stay the same.
Height = \(\frac{1}{4} + \frac{1}{4} = \frac{2}{4}\) m.
Volume \(V_2 = 50 \times \frac{1}{2} \times \frac{2}{4} = 12.50 \text{ m}^3\).
Step 3:
Height = \(\frac{3}{4}\) m.
Volume \(V_3 = 50 \times \frac{1}{2} \times \frac{3}{4} = 18.75 \text{ m}^3\).
This forms an AP: 6.25, 12.50, 18.75, ...
\(a = 6.25\), \(d = 6.25\), \(n = 15\).
Total Volume = \(S_{15} = \frac{15}{2}[2a + (15-1)d]\)
\[ S_{15} = \frac{15}{2}[2(6.25) + 14(6.25)] \]
\[ S_{15} = \frac{15}{2}[16 \times 6.25] \]
\[ S_{15} = 15 \times 8 \times 6.25 \]
\[ S_{15} = 120 \times 6.25 = 750 \text{ m}^3 \]
Answer: The total volume of concrete required is 750 m³.