Class 10 NCERT Solutions
Chapter 5: Arithmetic Progressions
Master the common difference, the $n^{th}$ term logic, and the summation of arithmetic series to decode mathematical patterns with our step-by-step guidance.
Exercise 5.1
1. In which of the following situations does the list of numbers involved make an arithmetic progression, and why?
(i) Taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
(ii) Amount of air in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining at a time.
(iii) Cost of digging a well after every metre when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.
(iv) Amount of money in the account every year when ₹10000 is deposited at compound interest at 8% per annum.
(i) Taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
(ii) Amount of air in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining at a time.
(iii) Cost of digging a well after every metre when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.
(iv) Amount of money in the account every year when ₹10000 is deposited at compound interest at 8% per annum.
(i) Yes. The fare sequence is ₹15, ₹23, ₹31, … which is an A.P. with a constant common difference of ₹8.
(ii) No. The sequence $1,\ \frac{3}{4},\ \frac{9}{16},\ \frac{27}{64},\ldots$ is not an A.P. because the difference between consecutive terms is not constant.
(iii) Yes. The cost sequence is ₹150, ₹200, ₹250, … which is an A.P. with a constant common difference of ₹50.
(iv) No. The sequence $₹10000\times\frac{108}{100},\ ₹10000\times\left(\frac{108}{100}\right)^2,\ldots$ is not an A.P. because the difference between consecutive terms keeps changing.
(ii) No. The sequence $1,\ \frac{3}{4},\ \frac{9}{16},\ \frac{27}{64},\ldots$ is not an A.P. because the difference between consecutive terms is not constant.
(iii) Yes. The cost sequence is ₹150, ₹200, ₹250, … which is an A.P. with a constant common difference of ₹50.
(iv) No. The sequence $₹10000\times\frac{108}{100},\ ₹10000\times\left(\frac{108}{100}\right)^2,\ldots$ is not an A.P. because the difference between consecutive terms keeps changing.
2. Write the first four terms of the A.P. when the first term $a$ and the common difference $d$ are given:
(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) $a=-1.25,\ d=-0.25$
Using the general form of an A.P.: $a,\ a+d,\ a+2d,\ a+3d,\ldots$
(i) $10,\ 20,\ 30,\ 40$
(ii) $-2,\ -2,\ -2,\ -2$
(iii) $4,\ 1,\ -2,\ -5$
(iv) $-1,\ -\frac{1}{2},\ 0,\ \frac{1}{2}$
(v) $-1.25,\ -1.50,\ -1.75,\ -2.00$
(i) $10,\ 20,\ 30,\ 40$
(ii) $-2,\ -2,\ -2,\ -2$
(iii) $4,\ 1,\ -2,\ -5$
(iv) $-1,\ -\frac{1}{2},\ 0,\ \frac{1}{2}$
(v) $-1.25,\ -1.50,\ -1.75,\ -2.00$
3. For the following A.P.s, write the first term and the common difference:
(i) $3,1,-1,-3,\ldots$ (ii) $-5,-1,3,7,\ldots$ (iii) $\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},\ldots$ (iv) $0.6,1.7,2.8,3.9,\ldots$
(i) $3,1,-1,-3,\ldots$ (ii) $-5,-1,3,7,\ldots$ (iii) $\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},\ldots$ (iv) $0.6,1.7,2.8,3.9,\ldots$
(i) $a=3,\ d=1-3=-2$
(ii) $a=-5,\ d=-1+5=4$
(iii) $a=\frac{1}{3},\ d=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}$
(iv) $a=0.6,\ d=1.7-0.6=1.1$
(ii) $a=-5,\ d=-1+5=4$
(iii) $a=\frac{1}{3},\ d=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}$
(iv) $a=0.6,\ d=1.7-0.6=1.1$
4. Which of the following are A.P.s? If they form an A.P., find the common difference $d$ and write three more terms:
(i) $2,4,8,16,\ldots$ (ii) $2,\frac{5}{2},3,\frac{7}{2},\ldots$ (iii) $-1.2,-3.2,-5.2,-7.2,\ldots$
(iv) $-10,-6,-2,2,\ldots$ (v) $3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},\ldots$ (vi) $0.2,0.22,0.222,0.2222,\ldots$
(vii) $0,-4,-8,-12,\ldots$ (viii) $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},\ldots$ (ix) $1,3,9,27,\ldots$
(x) $a,2a,3a,4a,\ldots$ (xi) $a,a^2,a^3,a^4,\ldots$ (xii) $\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},\ldots$
(xiii) $\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},\ldots$ (xiv) $1^2,3^2,5^2,7^2,\ldots$ (xv) $1^2,5^2,7^2,73,\ldots$
(i) $2,4,8,16,\ldots$ (ii) $2,\frac{5}{2},3,\frac{7}{2},\ldots$ (iii) $-1.2,-3.2,-5.2,-7.2,\ldots$
(iv) $-10,-6,-2,2,\ldots$ (v) $3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},\ldots$ (vi) $0.2,0.22,0.222,0.2222,\ldots$
(vii) $0,-4,-8,-12,\ldots$ (viii) $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},\ldots$ (ix) $1,3,9,27,\ldots$
(x) $a,2a,3a,4a,\ldots$ (xi) $a,a^2,a^3,a^4,\ldots$ (xii) $\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32},\ldots$
(xiii) $\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},\ldots$ (xiv) $1^2,3^2,5^2,7^2,\ldots$ (xv) $1^2,5^2,7^2,73,\ldots$
(i) Differences: $4-2=2;\ 8-4=4;\ 16-8=8$. Not constant. Not an A.P.
(ii) Differences: $\frac{5}{2}-2=\frac{1}{2};\ 3-\frac{5}{2}=\frac{1}{2};\ \frac{7}{2}-3=\frac{1}{2}$. Constant, $d=\frac{1}{2}$. It is an A.P.
Next three terms: $4,\ \frac{9}{2},\ 5$
(iii) Differences: $-3.2-(-1.2)=-2;\ -5.2-(-3.2)=-2;\ -7.2-(-5.2)=-2$. Constant, $d=-2$. It is an A.P.
Next three terms: $-9.2,\ -11.2,\ -13.2$
(iv) Differences: $-6-(-10)=4;\ -2-(-6)=4;\ 2-(-2)=4$. Constant, $d=4$. It is an A.P.
Next three terms: $6,\ 10,\ 14$
(v) Differences: $(3+\sqrt{2})-3=\sqrt{2};\ (3+2\sqrt{2})-(3+\sqrt{2})=\sqrt{2};\ (3+3\sqrt{2})-(3+2\sqrt{2})=\sqrt{2}$. Constant, $d=\sqrt{2}$. It is an A.P.
Next three 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 constant. Not an A.P.
(vii) Differences: $-4-0=-4;\ -8-(-4)=-4;\ -12-(-8)=-4$. Constant, $d=-4$. It is an A.P.
Next three terms: $-16,\ -20,\ -24$
(viii) All differences $= 0$. Constant, $d=0$. It is an A.P.
Next three terms: $-\frac{1}{2},\ -\frac{1}{2},\ -\frac{1}{2}$
(ix) $3-1=2;\ 9-3=6;\ 27-9=18$. Not constant. Not an A.P.
(x) $2a-a=a;\ 3a-2a=a;\ 4a-3a=a$. Constant, $d=a$. It is an A.P.
Next three terms: $5a,\ 6a,\ 7a$
(xi) $a^2-a=a(a-1);\ a^3-a^2=a^2(a-1)$. Not constant (unless $a=1$). Not an A.P.
(xii) $\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2};\ \sqrt{18}-\sqrt{8}=3\sqrt{2}-2\sqrt{2}=\sqrt{2};\ \sqrt{32}-\sqrt{18}=4\sqrt{2}-3\sqrt{2}=\sqrt{2}$. Constant, $d=\sqrt{2}$. It is an A.P.
Next three terms: $\sqrt{50},\ \sqrt{72},\ \sqrt{98}$
(xiii) $\sqrt{6}-\sqrt{3}\neq\sqrt{9}-\sqrt{6}$. Not constant. Not an A.P.
(xiv) The sequence is $1,9,25,49,\ldots$; differences: $8, 16, 24$. Not constant. Not an A.P.
(xv) The sequence is $1,25,49,73,\ldots$; differences: $24, 24, 24$. Constant, $d=24$. It is an A.P.
Next three terms: $97,\ 121,\ 145$
(ii) Differences: $\frac{5}{2}-2=\frac{1}{2};\ 3-\frac{5}{2}=\frac{1}{2};\ \frac{7}{2}-3=\frac{1}{2}$. Constant, $d=\frac{1}{2}$. It is an A.P.
Next three terms: $4,\ \frac{9}{2},\ 5$
(iii) Differences: $-3.2-(-1.2)=-2;\ -5.2-(-3.2)=-2;\ -7.2-(-5.2)=-2$. Constant, $d=-2$. It is an A.P.
Next three terms: $-9.2,\ -11.2,\ -13.2$
(iv) Differences: $-6-(-10)=4;\ -2-(-6)=4;\ 2-(-2)=4$. Constant, $d=4$. It is an A.P.
Next three terms: $6,\ 10,\ 14$
(v) Differences: $(3+\sqrt{2})-3=\sqrt{2};\ (3+2\sqrt{2})-(3+\sqrt{2})=\sqrt{2};\ (3+3\sqrt{2})-(3+2\sqrt{2})=\sqrt{2}$. Constant, $d=\sqrt{2}$. It is an A.P.
Next three 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 constant. Not an A.P.
(vii) Differences: $-4-0=-4;\ -8-(-4)=-4;\ -12-(-8)=-4$. Constant, $d=-4$. It is an A.P.
Next three terms: $-16,\ -20,\ -24$
(viii) All differences $= 0$. Constant, $d=0$. It is an A.P.
Next three terms: $-\frac{1}{2},\ -\frac{1}{2},\ -\frac{1}{2}$
(ix) $3-1=2;\ 9-3=6;\ 27-9=18$. Not constant. Not an A.P.
(x) $2a-a=a;\ 3a-2a=a;\ 4a-3a=a$. Constant, $d=a$. It is an A.P.
Next three terms: $5a,\ 6a,\ 7a$
(xi) $a^2-a=a(a-1);\ a^3-a^2=a^2(a-1)$. Not constant (unless $a=1$). Not an A.P.
(xii) $\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2};\ \sqrt{18}-\sqrt{8}=3\sqrt{2}-2\sqrt{2}=\sqrt{2};\ \sqrt{32}-\sqrt{18}=4\sqrt{2}-3\sqrt{2}=\sqrt{2}$. Constant, $d=\sqrt{2}$. It is an A.P.
Next three terms: $\sqrt{50},\ \sqrt{72},\ \sqrt{98}$
(xiii) $\sqrt{6}-\sqrt{3}\neq\sqrt{9}-\sqrt{6}$. Not constant. Not an A.P.
(xiv) The sequence is $1,9,25,49,\ldots$; differences: $8, 16, 24$. Not constant. Not an A.P.
(xv) The sequence is $1,25,49,73,\ldots$; differences: $24, 24, 24$. Constant, $d=24$. It is an A.P.
Next three terms: $97,\ 121,\ 145$
Exercise 5.2
1. Fill in the blanks given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the A.P.:
(i) $a=7,\ d=3,\ n=8,\ a_n=?$ (ii) $a=-18,\ d=?,\ n=10,\ a_n=0$
(iii) $a=?,\ d=-3,\ n=18,\ a_n=-5$ (iv) $a=-18.9,\ d=2.5,\ n=?,\ a_n=3.6$
(v) $a=3.5,\ d=0,\ n=105,\ a_n=?$
(i) $a=7,\ d=3,\ n=8,\ a_n=?$ (ii) $a=-18,\ d=?,\ n=10,\ a_n=0$
(iii) $a=?,\ d=-3,\ n=18,\ a_n=-5$ (iv) $a=-18.9,\ d=2.5,\ n=?,\ a_n=3.6$
(v) $a=3.5,\ d=0,\ n=105,\ a_n=?$
(i) $a_n=a+(n-1)d=7+(8-1)\times3=7+21=28$
(ii) $a_n=a+(n-1)d \Rightarrow 0=-18+(10-1)d \Rightarrow 9d=18 \Rightarrow d=2$
(iii) $a_n=a+(n-1)d \Rightarrow -5=a+(18-1)(-3) \Rightarrow a=-5+51=46$
(iv) $3.6=-18.9+(n-1)(2.5) \Rightarrow 22.5=2.5n-2.5 \Rightarrow 2.5n=25 \Rightarrow n=10$
(v) $a_n=a+(n-1)d=3.5+(105-1)\times0=3.5$
(ii) $a_n=a+(n-1)d \Rightarrow 0=-18+(10-1)d \Rightarrow 9d=18 \Rightarrow d=2$
(iii) $a_n=a+(n-1)d \Rightarrow -5=a+(18-1)(-3) \Rightarrow a=-5+51=46$
(iv) $3.6=-18.9+(n-1)(2.5) \Rightarrow 22.5=2.5n-2.5 \Rightarrow 2.5n=25 \Rightarrow n=10$
(v) $a_n=a+(n-1)d=3.5+(105-1)\times0=3.5$
2. Choose the correct option and justify:
(i) 30th term of the A.P.: $10,7,4,\ldots$ is: (A) 97 (B) 77 (C) $-77$ (D) $-87$
(ii) 11th term of the A.P.: $-3,-\frac{1}{2},2,\ldots$ is: (A) 28 (B) 22 (C) $-38$ (D) $-48\frac{1}{2}$
(i) 30th term of the A.P.: $10,7,4,\ldots$ is: (A) 97 (B) 77 (C) $-77$ (D) $-87$
(ii) 11th term of the A.P.: $-3,-\frac{1}{2},2,\ldots$ is: (A) 28 (B) 22 (C) $-38$ (D) $-48\frac{1}{2}$
(i) $a=10,\ d=7-10=-3$
$a_{30}=10+(30-1)(-3)=10-87=-77$
Option (C) is correct.
(ii) $a=-3,\ d=-\frac{1}{2}+3=\frac{5}{2}$
$a_{11}=-3+(11-1)\times\frac{5}{2}=-3+25=22$
Option (B) is correct.
$a_{30}=10+(30-1)(-3)=10-87=-77$
Option (C) is correct.
(ii) $a=-3,\ d=-\frac{1}{2}+3=\frac{5}{2}$
$a_{11}=-3+(11-1)\times\frac{5}{2}=-3+25=22$
Option (B) is correct.
3. In the following A.P.s, find the missing terms:
(i) $2,\ \square,\ 26$ (ii) $\square,\ 13,\ \square,\ 3$ (iii) $5,\ \square,\ \square,\ 9\frac{1}{2}$
(iv) $-4,\ \square,\ \square,\ \square,\ \square,\ 6$ (v) $\square,\ 38,\ \square,\ \square,\ \square,\ -22$
(i) $2,\ \square,\ 26$ (ii) $\square,\ 13,\ \square,\ 3$ (iii) $5,\ \square,\ \square,\ 9\frac{1}{2}$
(iv) $-4,\ \square,\ \square,\ \square,\ \square,\ 6$ (v) $\square,\ 38,\ \square,\ \square,\ \square,\ -22$
(i) $a_1=2,\ a_3=26 \Rightarrow 26=2+(3-1)d \Rightarrow 24=2d \Rightarrow d=12$
$a_2=2+12=14$. Missing term: 14
(ii) $a_2=13 \Rightarrow a+d=13$ and $a_4=3 \Rightarrow a+3d=3$
Solving: $a=18,\ d=-5$
$a_3=18+2(-5)=8$. Missing terms: 18 and 8
(iii) $a_1=5,\ a_4=\frac{19}{2} \Rightarrow \frac{19}{2}=5+3d \Rightarrow 3d=\frac{9}{2} \Rightarrow d=\frac{3}{2}$
$a_2=5+\frac{3}{2}=\frac{13}{2}=6\frac{1}{2}$ and $a_3=\frac{13}{2}+\frac{3}{2}=8$. Missing terms: $6\frac{1}{2}$ and $8$
(iv) $a=-4,\ a_6=6 \Rightarrow -4+5d=6 \Rightarrow d=2$
$a_2=-2,\ a_3=0,\ a_4=2,\ a_5=4$. Missing terms: $-2, 0, 2, 4$
(v) $a+d=38$ and $a+5d=-22$. Solving: $a=53,\ d=-15$
$a_3=53+2(-15)=23,\ a_4=8,\ a_5=-7$. Missing terms (1st, 3rd, 4th, 5th): $53,\ 23,\ 8,\ -7$
$a_2=2+12=14$. Missing term: 14
(ii) $a_2=13 \Rightarrow a+d=13$ and $a_4=3 \Rightarrow a+3d=3$
Solving: $a=18,\ d=-5$
$a_3=18+2(-5)=8$. Missing terms: 18 and 8
(iii) $a_1=5,\ a_4=\frac{19}{2} \Rightarrow \frac{19}{2}=5+3d \Rightarrow 3d=\frac{9}{2} \Rightarrow d=\frac{3}{2}$
$a_2=5+\frac{3}{2}=\frac{13}{2}=6\frac{1}{2}$ and $a_3=\frac{13}{2}+\frac{3}{2}=8$. Missing terms: $6\frac{1}{2}$ and $8$
(iv) $a=-4,\ a_6=6 \Rightarrow -4+5d=6 \Rightarrow d=2$
$a_2=-2,\ a_3=0,\ a_4=2,\ a_5=4$. Missing terms: $-2, 0, 2, 4$
(v) $a+d=38$ and $a+5d=-22$. Solving: $a=53,\ d=-15$
$a_3=53+2(-15)=23,\ a_4=8,\ a_5=-7$. Missing terms (1st, 3rd, 4th, 5th): $53,\ 23,\ 8,\ -7$
4. Which term of the A.P.: $3,8,13,18,\ldots$ is 78?
Let $a_n=78$, where $a=3$ and $d=5$.
$78=3+(n-1)\times5 \Rightarrow 75=(n-1)\times5 \Rightarrow n-1=15 \Rightarrow n=16$
Therefore, 78 is the 16th term of the A.P.
$78=3+(n-1)\times5 \Rightarrow 75=(n-1)\times5 \Rightarrow n-1=15 \Rightarrow n=16$
Therefore, 78 is the 16th term of the A.P.
5. Find the number of terms in each of the following A.P.s:
(i) $7,13,19,\ldots,205$ (ii) $18,15\frac{1}{2},13,\ldots,-47$
(i) $7,13,19,\ldots,205$ (ii) $18,15\frac{1}{2},13,\ldots,-47$
(i) $a=7,\ d=6,\ a_n=205$
$205=7+(n-1)\times6 \Rightarrow 198=(n-1)\times6 \Rightarrow n-1=33 \Rightarrow n=34$
There are 34 terms.
(ii) $a=18,\ d=-\frac{5}{2},\ a_n=-47$
$-47=18+(n-1)\left(-\frac{5}{2}\right) \Rightarrow -65=(n-1)\left(-\frac{5}{2}\right) \Rightarrow n-1=26 \Rightarrow n=27$
There are 27 terms.
$205=7+(n-1)\times6 \Rightarrow 198=(n-1)\times6 \Rightarrow n-1=33 \Rightarrow n=34$
There are 34 terms.
(ii) $a=18,\ d=-\frac{5}{2},\ a_n=-47$
$-47=18+(n-1)\left(-\frac{5}{2}\right) \Rightarrow -65=(n-1)\left(-\frac{5}{2}\right) \Rightarrow n-1=26 \Rightarrow n=27$
There are 27 terms.
6. Check whether $-150$ is a term of the A.P.: $11,8,5,2,\ldots$
Let $a_n=-150$, where $a=11$ and $d=-3$.
$-150=11+(n-1)(-3) \Rightarrow -161=-3n+3 \Rightarrow 3n=164 \Rightarrow n=\frac{164}{3}=54.67$
Since $n$ is not a natural number, $-150$ is not a term of the given A.P.
$-150=11+(n-1)(-3) \Rightarrow -161=-3n+3 \Rightarrow 3n=164 \Rightarrow n=\frac{164}{3}=54.67$
Since $n$ is not a natural number, $-150$ is not a term of the given A.P.
7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
$a+10d=38$ and $a+15d=73$
Subtracting: $5d=35 \Rightarrow d=7$, then $a=-32$
$a_{31}=-32+30\times7=-32+210=178$
The 31st term is 178.
Subtracting: $5d=35 \Rightarrow d=7$, then $a=-32$
$a_{31}=-32+30\times7=-32+210=178$
The 31st term is 178.
8. An A.P. consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.
$a_3=12 \Rightarrow a+2d=12$ and $a_{50}=106 \Rightarrow a+49d=106$
Solving both equations: $a=8,\ d=2$
$a_{29}=8+(29-1)\times2=8+56=64$
The 29th term is 64.
Solving both equations: $a=8,\ d=2$
$a_{29}=8+(29-1)\times2=8+56=64$
The 29th term is 64.
9. If the 3rd and the 9th terms of an A.P. are 4 and $-8$ respectively, which term of this A.P. is zero?
$a+2d=4$ and $a+8d=-8$
Solving: $a=8,\ d=-2$
Let $a_n=0$: $8+(n-1)(-2)=0 \Rightarrow 8-2n+2=0 \Rightarrow n=5$
The 5th term of the A.P. is zero.
Solving: $a=8,\ d=-2$
Let $a_n=0$: $8+(n-1)(-2)=0 \Rightarrow 8-2n+2=0 \Rightarrow n=5$
The 5th term of the A.P. is zero.
10. The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
$a_{17}=a_{10}+7$
$\Rightarrow a+16d=a+9d+7$
$\Rightarrow 7d=7 \Rightarrow d=1$
The common difference is $d=1$.
$\Rightarrow a+16d=a+9d+7$
$\Rightarrow 7d=7 \Rightarrow d=1$
The common difference is $d=1$.
11. Which term of the A.P.: $3,15,27,39,\ldots$ will be 132 more than its 54th term?
Here $a=3,\ d=12$. Let the $n$th term be 132 more than the 54th term.
$a_n=a_{54}+132$
$3+(n-1)\times12=3+(54-1)\times12+132$
$\Rightarrow (n-1)\times12=768 \Rightarrow n-1=64 \Rightarrow n=65$
The 65th term is 132 more than the 54th term.
$a_n=a_{54}+132$
$3+(n-1)\times12=3+(54-1)\times12+132$
$\Rightarrow (n-1)\times12=768 \Rightarrow n-1=64 \Rightarrow n=65$
The 65th term is 132 more than the 54th term.
12. Two A.P.s have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms?
Let the common difference be $d$ and the two A.P.s be $\{a_n\}$ and $\{a_n’\}$.
$a_{100}-a_{100}’=(a_1+99d)-(a_1’+99d)=a_1-a_1’=100$
Now: $a_{1000}-a_{1000}’=(a_1+999d)-(a_1’+999d)=a_1-a_1’=100$
The difference between their 1000th terms is also 100.
$a_{100}-a_{100}’=(a_1+99d)-(a_1’+99d)=a_1-a_1’=100$
Now: $a_{1000}-a_{1000}’=(a_1+999d)-(a_1’+999d)=a_1-a_1’=100$
The difference between their 1000th terms is also 100.
13. How many three-digit numbers are divisible by 7?
Three-digit numbers divisible by 7 form the sequence $105, 112, \ldots, 994$.
Let the total count be $n$: $994=105+(n-1)\times7 \Rightarrow 889=(n-1)\times7 \Rightarrow n-1=127 \Rightarrow n=128$
There are 128 three-digit numbers divisible by 7.
Let the total count be $n$: $994=105+(n-1)\times7 \Rightarrow 889=(n-1)\times7 \Rightarrow n-1=127 \Rightarrow n=128$
There are 128 three-digit numbers divisible by 7.
14. How many multiples of 4 lie between 10 and 250?
Multiples of 4 between 10 and 250: $12, 16, \ldots, 248$.
Here $a=12,\ d=4,\ a_n=248$.
$248=12+(n-1)\times4 \Rightarrow 236=(n-1)\times4 \Rightarrow n-1=59 \Rightarrow n=60$
There are 60 multiples of 4 between 10 and 250.
Here $a=12,\ d=4,\ a_n=248$.
$248=12+(n-1)\times4 \Rightarrow 236=(n-1)\times4 \Rightarrow n-1=59 \Rightarrow n=60$
There are 60 multiples of 4 between 10 and 250.
15. For what value of $n$, are the $n$th terms of two A.P.s: $63,65,67,\ldots$ and $3,10,17,\ldots$ equal?
Set the $n$th terms equal:
$63+(n-1)\times2=3+(n-1)\times7$
$\Rightarrow 60=5(n-1) \Rightarrow n-1=12 \Rightarrow n=13$
The $n$th terms of the two A.P.s are equal when $n=\mathbf{13}$.
$63+(n-1)\times2=3+(n-1)\times7$
$\Rightarrow 60=5(n-1) \Rightarrow n-1=12 \Rightarrow n=13$
The $n$th terms of the two A.P.s are equal when $n=\mathbf{13}$.
16. Determine the A.P. whose third term is 16 and 7th term exceeds the 5th term by 12.
$a_3=16 \Rightarrow a+2d=16$ and $a_7=a_5+12 \Rightarrow a+6d=a+4d+12 \Rightarrow 2d=12 \Rightarrow d=6$
Then $a+12=16 \Rightarrow a=4$
The A.P. is $4, 10, 16, 22, \ldots$
Then $a+12=16 \Rightarrow a=4$
The A.P. is $4, 10, 16, 22, \ldots$
17. Find the 20th term from the last term of the A.P.: $3,8,13,\ldots,253$.
Counting from the end, treat $a=253$ and $d=-5$.
20th term from the end $= 253+(20-1)(-5)=253-95=158$
The 20th term from the last is 158.
20th term from the end $= 253+(20-1)(-5)=253-95=158$
The 20th term from the last is 158.
18. The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
$(a+3d)+(a+7d)=24 \Rightarrow 2a+10d=24 \Rightarrow a+5d=12\ \ldots(i)$
$(a+5d)+(a+9d)=44 \Rightarrow 2a+14d=44 \Rightarrow a+7d=22\ \ldots(ii)$
Solving (i) and (ii): $a=-13,\ d=5$
The first three terms are $-13,\ -8,\ -3$.
$(a+5d)+(a+9d)=44 \Rightarrow 2a+14d=44 \Rightarrow a+7d=22\ \ldots(ii)$
Solving (i) and (ii): $a=-13,\ d=5$
The first three terms are $-13,\ -8,\ -3$.
19. 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?
Let $a_n=7000$, where $a=5000$ and $d=200$.
$7000=5000+(n-1)\times200 \Rightarrow 2000=(n-1)\times200 \Rightarrow n-1=10 \Rightarrow n=11$
The 11th year corresponds to 2005 (1995 + 10 = 2005; income reaches ₹7000 in 2005).
$7000=5000+(n-1)\times200 \Rightarrow 2000=(n-1)\times200 \Rightarrow n-1=10 \Rightarrow n=11$
The 11th year corresponds to 2005 (1995 + 10 = 2005; income reaches ₹7000 in 2005).
20. Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the $n$th week her weekly savings become ₹20.75, find $n$.
$a_n=20.75$, where $a=5$ and $d=1.75$.
$20.75=5+(n-1)\times1.75 \Rightarrow 15.75=(n-1)\times1.75 \Rightarrow n-1=9 \Rightarrow n=10$
In the 10th week, her savings will be ₹20.75.
$20.75=5+(n-1)\times1.75 \Rightarrow 15.75=(n-1)\times1.75 \Rightarrow n-1=9 \Rightarrow n=10$
In the 10th week, her savings will be ₹20.75.
Exercise 5.3
1. Find the sum of the following A.P.s:
(i) $2,7,12,\ldots$ to 10 terms (ii) $-37,-33,-29,\ldots$ to 12 terms
(iii) $0.6,1.7,2.8,\ldots$ to 100 terms (iv) $\frac{1}{15},\frac{1}{12},\frac{1}{10},\ldots$ to 11 terms
(i) $2,7,12,\ldots$ to 10 terms (ii) $-37,-33,-29,\ldots$ to 12 terms
(iii) $0.6,1.7,2.8,\ldots$ to 100 terms (iv) $\frac{1}{15},\frac{1}{12},\frac{1}{10},\ldots$ to 11 terms
(i) $a=2,\ d=5,\ n=10$
$S_{10}=\frac{10}{2}[2\times2+(10-1)\times5]=5\times[4+45]=5\times49=245$
(ii) $a=-37,\ d=4,\ n=12$
$S_{12}=\frac{12}{2}[2\times(-37)+(12-1)\times4]=6\times[-74+44]=6\times(-30)=-180$
(iii) $a=0.6,\ d=1.1,\ n=100$
$S_{100}=\frac{100}{2}[2\times0.6+(100-1)\times1.1]=50\times[1.2+108.9]=50\times110.1=5505$
(iv) $a=\frac{1}{15},\ d=\frac{1}{12}-\frac{1}{15}=\frac{1}{60},\ n=11$
$S_{11}=\frac{11}{2}\left[\frac{2}{15}+(11-1)\times\frac{1}{60}\right]=\frac{11}{2}\left[\frac{2}{15}+\frac{1}{6}\right]=\frac{11}{2}\times\frac{3}{10}=\frac{33}{20}$
$S_{10}=\frac{10}{2}[2\times2+(10-1)\times5]=5\times[4+45]=5\times49=245$
(ii) $a=-37,\ d=4,\ n=12$
$S_{12}=\frac{12}{2}[2\times(-37)+(12-1)\times4]=6\times[-74+44]=6\times(-30)=-180$
(iii) $a=0.6,\ d=1.1,\ n=100$
$S_{100}=\frac{100}{2}[2\times0.6+(100-1)\times1.1]=50\times[1.2+108.9]=50\times110.1=5505$
(iv) $a=\frac{1}{15},\ d=\frac{1}{12}-\frac{1}{15}=\frac{1}{60},\ n=11$
$S_{11}=\frac{11}{2}\left[\frac{2}{15}+(11-1)\times\frac{1}{60}\right]=\frac{11}{2}\left[\frac{2}{15}+\frac{1}{6}\right]=\frac{11}{2}\times\frac{3}{10}=\frac{33}{20}$
2. Find the sums given below:
(i) $7+10\frac{1}{2}+14+\ldots+84$ (ii) $34+32+30+\ldots+10$ (iii) $-5+(-8)+(-11)+\ldots+(-230)$
(i) $7+10\frac{1}{2}+14+\ldots+84$ (ii) $34+32+30+\ldots+10$ (iii) $-5+(-8)+(-11)+\ldots+(-230)$
(i) $a=7,\ d=\frac{7}{2},\ l=84$
$84=7+(n-1)\frac{7}{2} \Rightarrow 77=(n-1)\frac{7}{2} \Rightarrow n=23$
$S_{23}=\frac{23}{2}(7+84)=\frac{23}{2}\times91=\frac{2093}{2}=1046\frac{1}{2}$
(ii) $a=34,\ d=-2,\ a_n=10$
$10=34+(n-1)(-2) \Rightarrow -24=(n-1)(-2) \Rightarrow n=13$
$S_{13}=\frac{13}{2}(34+10)=\frac{13}{2}\times44=286$
(iii) $a=-5,\ d=-3,\ l=-230$
$-230=-5+(n-1)(-3) \Rightarrow -225=(n-1)(-3) \Rightarrow n=76$
$S_{76}=\frac{76}{2}(-5-230)=38\times(-235)=-8930$
$84=7+(n-1)\frac{7}{2} \Rightarrow 77=(n-1)\frac{7}{2} \Rightarrow n=23$
$S_{23}=\frac{23}{2}(7+84)=\frac{23}{2}\times91=\frac{2093}{2}=1046\frac{1}{2}$
(ii) $a=34,\ d=-2,\ a_n=10$
$10=34+(n-1)(-2) \Rightarrow -24=(n-1)(-2) \Rightarrow n=13$
$S_{13}=\frac{13}{2}(34+10)=\frac{13}{2}\times44=286$
(iii) $a=-5,\ d=-3,\ l=-230$
$-230=-5+(n-1)(-3) \Rightarrow -225=(n-1)(-3) \Rightarrow n=76$
$S_{76}=\frac{76}{2}(-5-230)=38\times(-235)=-8930$
3. In an A.P.:
(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 $a$.
(i) $50=5+(n-1)\times3 \Rightarrow 45=(n-1)\times3 \Rightarrow n=16$
$S_{16}=\frac{16}{2}(2\times5+15\times3)=8\times55=440$
(ii) $35=7+12d \Rightarrow d=\frac{7}{3}$
$S_{13}=\frac{13}{2}\left[14+12\times\frac{7}{3}\right]=\frac{13}{2}\times42=273$
(iii) $a_{12}=37 \Rightarrow a+11\times3=37 \Rightarrow a=4$
$S_{12}=\frac{12}{2}[2\times4+(12-1)\times3]=6\times41=246$
(iv) $a+2d=15$ and $125=\frac{10}{2}[2a+9d] \Rightarrow 25=2a+9d$
Solving: $a=17,\ d=-1$
$a_{10}=17+9\times(-1)=8$
(v) $S_9=75 \Rightarrow 75=\frac{9}{2}[2a+8\times5] \Rightarrow a=-\frac{35}{3}$
$a_9=-\frac{35}{3}+8\times5=\frac{-35+120}{3}=\frac{85}{3}$
(vi) $90=\frac{n}{2}[4+(n-1)\times8]=\frac{n}{2}[8n-4]=4n^2-2n$
$\Rightarrow 2n^2-n-45=0 \Rightarrow (2n+9)(n-5)=0$
$\Rightarrow n=5$ (rejecting $n=-\frac{9}{2}$)
$a_5=2+4\times8=34$
(vii) $210=\frac{n}{2}(8+62)=\frac{n}{2}\times70 \Rightarrow n=6$
$a_6=62 \Rightarrow 62=8+5d \Rightarrow d=\frac{54}{5}$
(viii) $a+(n-1)\times2=4\ \ldots(i)$
$-14=\frac{n}{2}[2a+(n-1)\times2] \Rightarrow -14=n(5-n) \Rightarrow n^2-5n-14=0$
$\Rightarrow (n-7)(n+2)=0 \Rightarrow n=7$ (rejecting $n=-2$)
From (i): $a+12=4 \Rightarrow a=-8$
(ix) $192=\frac{8}{2}[6+7d]=4[6+7d] \Rightarrow 48=6+7d \Rightarrow d=6$
(x) $S_9=\frac{9}{2}(a+28)=144 \Rightarrow a+28=32 \Rightarrow a=4$
$S_{16}=\frac{16}{2}(2\times5+15\times3)=8\times55=440$
(ii) $35=7+12d \Rightarrow d=\frac{7}{3}$
$S_{13}=\frac{13}{2}\left[14+12\times\frac{7}{3}\right]=\frac{13}{2}\times42=273$
(iii) $a_{12}=37 \Rightarrow a+11\times3=37 \Rightarrow a=4$
$S_{12}=\frac{12}{2}[2\times4+(12-1)\times3]=6\times41=246$
(iv) $a+2d=15$ and $125=\frac{10}{2}[2a+9d] \Rightarrow 25=2a+9d$
Solving: $a=17,\ d=-1$
$a_{10}=17+9\times(-1)=8$
(v) $S_9=75 \Rightarrow 75=\frac{9}{2}[2a+8\times5] \Rightarrow a=-\frac{35}{3}$
$a_9=-\frac{35}{3}+8\times5=\frac{-35+120}{3}=\frac{85}{3}$
(vi) $90=\frac{n}{2}[4+(n-1)\times8]=\frac{n}{2}[8n-4]=4n^2-2n$
$\Rightarrow 2n^2-n-45=0 \Rightarrow (2n+9)(n-5)=0$
$\Rightarrow n=5$ (rejecting $n=-\frac{9}{2}$)
$a_5=2+4\times8=34$
(vii) $210=\frac{n}{2}(8+62)=\frac{n}{2}\times70 \Rightarrow n=6$
$a_6=62 \Rightarrow 62=8+5d \Rightarrow d=\frac{54}{5}$
(viii) $a+(n-1)\times2=4\ \ldots(i)$
$-14=\frac{n}{2}[2a+(n-1)\times2] \Rightarrow -14=n(5-n) \Rightarrow n^2-5n-14=0$
$\Rightarrow (n-7)(n+2)=0 \Rightarrow n=7$ (rejecting $n=-2$)
From (i): $a+12=4 \Rightarrow a=-8$
(ix) $192=\frac{8}{2}[6+7d]=4[6+7d] \Rightarrow 48=6+7d \Rightarrow d=6$
(x) $S_9=\frac{9}{2}(a+28)=144 \Rightarrow a+28=32 \Rightarrow a=4$
4. How many terms of the A.P.: $9,17,25,\ldots$ must be taken to give a sum of 636?
Let $S_n=636$, where $a=9$ and $d=8$.
$636=\frac{n}{2}[18+(n-1)\times8]=n[9+4n-4]=4n^2+5n$
$\Rightarrow 4n^2+5n-636=0$
$\Rightarrow 4n^2+53n-48n-636=0$
$\Rightarrow n(4n+53)-12(4n+53)=0$
$\Rightarrow (4n+53)(n-12)=0$
$\Rightarrow n=12$ (rejecting $n=-\frac{53}{4}$)
12 terms must be taken.
$636=\frac{n}{2}[18+(n-1)\times8]=n[9+4n-4]=4n^2+5n$
$\Rightarrow 4n^2+5n-636=0$
$\Rightarrow 4n^2+53n-48n-636=0$
$\Rightarrow n(4n+53)-12(4n+53)=0$
$\Rightarrow (4n+53)(n-12)=0$
$\Rightarrow n=12$ (rejecting $n=-\frac{53}{4}$)
12 terms must be taken.
5. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
$400=\frac{n}{2}(5+45)=\frac{n}{2}\times50=25n \Rightarrow n=16$
$a_{16}=45 \Rightarrow 5+15d=45 \Rightarrow d=\frac{40}{15}=\frac{8}{3}$
There are 16 terms and the common difference is $d=\frac{8}{3}$.
$a_{16}=45 \Rightarrow 5+15d=45 \Rightarrow d=\frac{40}{15}=\frac{8}{3}$
There are 16 terms and the common difference is $d=\frac{8}{3}$.
6. The first and last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
$a=17,\ a_n=350,\ d=9$
$350=17+(n-1)\times9 \Rightarrow 333=(n-1)\times9 \Rightarrow n=38$
$S_{38}=\frac{38}{2}(17+350)=19\times367=6973$
There are 38 terms and their sum is 6973.
$350=17+(n-1)\times9 \Rightarrow 333=(n-1)\times9 \Rightarrow n=38$
$S_{38}=\frac{38}{2}(17+350)=19\times367=6973$
There are 38 terms and their sum is 6973.
7. Find the sum of first 22 terms of an A.P. in which $d=7$ and the 22nd term is 149.
$a_{22}=149 \Rightarrow a+21\times7=149 \Rightarrow a=149-147=2$
$S_{22}=\frac{22}{2}[2\times2+(22-1)\times7]=11\times[4+147]=11\times151=1661$
The sum of the first 22 terms is 1661.
$S_{22}=\frac{22}{2}[2\times2+(22-1)\times7]=11\times[4+147]=11\times151=1661$
The sum of the first 22 terms is 1661.
8. Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
$a_2=14 \Rightarrow a+d=14$ and $a_3=18 \Rightarrow a+2d=18$
Solving: $a=10,\ d=4$
$S_{51}=\frac{51}{2}[2\times10+(51-1)\times4]=\frac{51}{2}[20+200]=\frac{51}{2}\times220=51\times110=5610$
The sum of the first 51 terms is 5610.
Solving: $a=10,\ d=4$
$S_{51}=\frac{51}{2}[2\times10+(51-1)\times4]=\frac{51}{2}[20+200]=\frac{51}{2}\times220=51\times110=5610$
The sum of the first 51 terms is 5610.
9. If the sum of first 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of first $n$ terms.
$\frac{7}{2}[2a+6d]=49 \Rightarrow a+3d=7\ \ldots(i)$
$\frac{17}{2}[2a+16d]=289 \Rightarrow a+8d=17\ \ldots(ii)$
Solving (i) and (ii): $a=1,\ d=2$
$S_n=\frac{n}{2}[2\times1+(n-1)\times2]=\frac{n}{2}\times2n=n^2$
The sum of first $n$ terms is $\mathbf{n^2}$.
$\frac{17}{2}[2a+16d]=289 \Rightarrow a+8d=17\ \ldots(ii)$
Solving (i) and (ii): $a=1,\ d=2$
$S_n=\frac{n}{2}[2\times1+(n-1)\times2]=\frac{n}{2}\times2n=n^2$
The sum of first $n$ terms is $\mathbf{n^2}$.
10. Show that $a_1,a_2,\ldots,a_n,\ldots$ form an A.P. where $a_n$ is defined as below. Also find the sum of the first 15 terms in each case:
(i) $a_n=3+4n$ (ii) $a_n=9-5n$
(i) $a_n=3+4n$ (ii) $a_n=9-5n$
(i) $d=a_n-a_{n-1}=(3+4n)-[3+4(n-1)]=4$ (constant).
Since $d$ is constant, the sequence is an A.P.
$a_1=7$
$S_{15}=\frac{15}{2}[2\times7+(15-1)\times4]=\frac{15}{2}\times70=525$
(ii) $d=(9-5n)-[9-5(n-1)]=-5$ (constant).
Since $d$ is constant, the sequence is an A.P.
$a_1=4$
$S_{15}=\frac{15}{2}[2\times4+(15-1)\times(-5)]=\frac{15}{2}\times(-62)=-465$
Since $d$ is constant, the sequence is an A.P.
$a_1=7$
$S_{15}=\frac{15}{2}[2\times7+(15-1)\times4]=\frac{15}{2}\times70=525$
(ii) $d=(9-5n)-[9-5(n-1)]=-5$ (constant).
Since $d$ is constant, the sequence is an A.P.
$a_1=4$
$S_{15}=\frac{15}{2}[2\times4+(15-1)\times(-5)]=\frac{15}{2}\times(-62)=-465$
11. If the sum of the first $n$ terms of an A.P. is $4n-n^2$, what is the first term? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the $n$th terms.
$S_n=4n-n^2$
$S_1=4(1)-(1)^2=3=a_1$ (first term)
$S_2=4(2)-(2)^2=4$ (sum of first two terms)
$a_n=S_n-S_{n-1}=(4n-n^2)-[4(n-1)-(n-1)^2]$
$=4n-n^2-4n+4+n^2-2n+1=5-2n$
$a_2=5-4=1,\quad a_3=5-6=-1,\quad a_{10}=5-20=-15$
The $n$th term is $a_n=5-2n$.
$S_1=4(1)-(1)^2=3=a_1$ (first term)
$S_2=4(2)-(2)^2=4$ (sum of first two terms)
$a_n=S_n-S_{n-1}=(4n-n^2)-[4(n-1)-(n-1)^2]$
$=4n-n^2-4n+4+n^2-2n+1=5-2n$
$a_2=5-4=1,\quad a_3=5-6=-1,\quad a_{10}=5-20=-15$
The $n$th term is $a_n=5-2n$.
12. Find the sum of the first 40 positive integers divisible by 6.
The sequence is $6, 12, 18, \ldots$ up to 40 terms.
$S_{40}=\frac{40}{2}[2\times6+(40-1)\times6]=20\times[12+234]=20\times246=4920$
The required sum is 4920.
$S_{40}=\frac{40}{2}[2\times6+(40-1)\times6]=20\times[12+234]=20\times246=4920$
The required sum is 4920.
13. Find the sum of the first 15 multiples of 8.
The sequence is $8, 16, 24, \ldots$ up to 15 terms.
$S_{15}=\frac{15}{2}[2\times8+(15-1)\times8]=\frac{15}{2}\times[16+112]=\frac{15}{2}\times128=960$
The required sum is 960.
$S_{15}=\frac{15}{2}[2\times8+(15-1)\times8]=\frac{15}{2}\times[16+112]=\frac{15}{2}\times128=960$
The required sum is 960.
14. Find the sum of the odd numbers between 0 and 50.
The odd numbers between 0 and 50 are $1, 3, 5, \ldots, 49$.
Here $n=25,\ a=1,\ d=2$.
$S_{25}=\frac{25}{2}[2\times1+(25-1)\times2]=\frac{25}{2}\times50=625$
The required sum is 625.
Here $n=25,\ a=1,\ d=2$.
$S_{25}=\frac{25}{2}[2\times1+(25-1)\times2]=\frac{25}{2}\times50=625$
The required sum is 625.
15. A contract on a construction job specifies a penalty for delay: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, with each succeeding day’s penalty ₹50 more than the preceding day’s. How much must the contractor pay if he delayed the work by 30 days?
The penalty sequence is $200, 250, 300, \ldots$ for 30 terms, with $a=200$ and $d=50$.
$S_{30}=\frac{30}{2}[2\times200+(30-1)\times50]=15\times[400+1450]=15\times1850=₹27750$
The contractor must pay a total penalty of ₹27750.
$S_{30}=\frac{30}{2}[2\times200+(30-1)\times50]=15\times[400+1450]=15\times1850=₹27750$
The contractor must pay a total penalty of ₹27750.
16. A sum of ₹700 is to be used to give seven cash prizes to students. If each prize is ₹20 less than its preceding prize, find the value of each prize.
Let the prizes be $a,\ a-20,\ a-40,\ldots$ for 7 terms, with $d=-20$.
$\frac{7}{2}[2a+(7-1)(-20)]=700$
$\Rightarrow \frac{7}{2}[2a-120]=700 \Rightarrow 2a-120=200 \Rightarrow a=160$
The prizes are: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60 and ₹40.
$\frac{7}{2}[2a+(7-1)(-20)]=700$
$\Rightarrow \frac{7}{2}[2a-120]=700 \Rightarrow 2a-120=200 \Rightarrow a=160$
The prizes are: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60 and ₹40.
17. In a school, students decided to plant trees in and around the school. Each section of each class will plant trees equal to the class number (Class I plants 1, Class II plants 2, etc., up to Class XII). There are three sections per class. How many trees will be planted?
Total trees planted $= 3+6+9+\ldots+36$ (12 terms, since $3\times12=36$).
$S_{12}=\frac{12}{2}[3+36]=6\times39=234$
A total of 234 trees will be planted.
$S_{12}=\frac{12}{2}[3+36]=6\times39=234$
A total of 234 trees will be planted.
18. A spiral is made up of successive semicircles with centres alternately at $A$ and $B$, of radii $0.5\ \text{cm},\ 1.0\ \text{cm},\ 1.5\ \text{cm},\ 2.0\ \text{cm},\ldots$ What is the total length of such a spiral made up of thirteen consecutive semicircles? $\left(\text{Take}\ \pi=\frac{22}{7}\right)$
The lengths of the semicircles form the A.P.: $0.5\pi,\ \pi,\ 1.5\pi,\ldots$ with $a=0.5\pi$ and $d=0.5\pi$.
$S_{13}=\frac{13}{2}[2\times0.5\pi+(13-1)\times0.5\pi]=\frac{13}{2}[\pi+6\pi]=\frac{13}{2}\times7\pi=\frac{13}{2}\times7\times\frac{22}{7}=143\ \text{cm}$
The total length of the spiral is 143 cm.
$S_{13}=\frac{13}{2}[2\times0.5\pi+(13-1)\times0.5\pi]=\frac{13}{2}[\pi+6\pi]=\frac{13}{2}\times7\pi=\frac{13}{2}\times7\times\frac{22}{7}=143\ \text{cm}$
The total length of the spiral is 143 cm.
19. 200 logs are stacked with 20 logs in the bottom row, 19 in the next, 18 in the next, and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
The sequence is $20, 19, 18, \ldots$ with $a=20$ and $d=-1$.
$\frac{n}{2}[40+(n-1)(-1)]=200 \Rightarrow \frac{n}{2}[41-n]=200$
$\Rightarrow 41n-n^2=400 \Rightarrow n^2-41n+400=0 \Rightarrow (n-25)(n-16)=0$
$\Rightarrow n=25$ or $n=16$
Checking: $a_{25}=20+24\times(-1)=-4<0$, which is impossible. So $n=25$ is rejected.
Therefore $n=16$ and $a_{16}=20+15\times(-1)=5$.
There are 16 rows and the top row has 5 logs.
$\frac{n}{2}[40+(n-1)(-1)]=200 \Rightarrow \frac{n}{2}[41-n]=200$
$\Rightarrow 41n-n^2=400 \Rightarrow n^2-41n+400=0 \Rightarrow (n-25)(n-16)=0$
$\Rightarrow n=25$ or $n=16$
Checking: $a_{25}=20+24\times(-1)=-4<0$, which is impossible. So $n=25$ is rejected.
Therefore $n=16$ and $a_{16}=20+15\times(-1)=5$.
There are 16 rows and the top row has 5 logs.
20. In a potato race, a bucket is placed 5 m from the first potato, with each subsequent potato 3 m further apart. There are ten potatoes. A competitor runs back and forth picking up one potato at a time. What is the total distance she has to run?
The distances for picking up successive potatoes form an A.P.: $10, 16, 22, \ldots$ with $a=10$ and $d=6$.
$S_{10}=\frac{10}{2}[2\times10+(10-1)\times6]=5\times[20+54]=5\times74=370\ \text{m}$
The total distance the competitor has to run is 370 m.
$S_{10}=\frac{10}{2}[2\times10+(10-1)\times6]=5\times[20+54]=5\times74=370\ \text{m}$
The total distance the competitor has to run is 370 m.
Exercise 5.4
1. Which term of the A.P.: $121,117,113,\ldots$ is its first negative term?
We need $a_n<0$, where $a=121$ and $d=-4$.
$121+(n-1)(-4)<0 \Rightarrow 125-4n<0 \Rightarrow 4n>125 \Rightarrow n>31.25$
The smallest integer greater than 31.25 is 32.
The 32nd term is the first negative term.
$121+(n-1)(-4)<0 \Rightarrow 125-4n<0 \Rightarrow 4n>125 \Rightarrow n>31.25$
The smallest integer greater than 31.25 is 32.
The 32nd term is the first negative term.
2. The sum of the third and the seventh terms of an A.P. is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.
$a_3+a_7=6 \Rightarrow (a+2d)+(a+6d)=6 \Rightarrow 2a+8d=6 \Rightarrow a+4d=3\ \ldots(i)$
$(a+2d)(a+6d)=8$
Using (i): $(3-2d)(3+2d)=8 \Rightarrow 9-4d^2=8 \Rightarrow d^2=\frac{1}{4} \Rightarrow d=\pm\frac{1}{2}$
When $d=\frac{1}{2}$: $a=1$
$S_{16}=\frac{16}{2}[2\times1+(16-1)\times\frac{1}{2}]=8\times[2+\frac{15}{2}]=8\times\frac{19}{2}=76$
When $d=-\frac{1}{2}$: $a=5$
$S_{16}=\frac{16}{2}[2\times5+(16-1)\times(-\frac{1}{2})]=8\times[10-\frac{15}{2}]=8\times\frac{5}{2}=20$
The sum of the first 16 terms is 76 or 20.
$(a+2d)(a+6d)=8$
Using (i): $(3-2d)(3+2d)=8 \Rightarrow 9-4d^2=8 \Rightarrow d^2=\frac{1}{4} \Rightarrow d=\pm\frac{1}{2}$
When $d=\frac{1}{2}$: $a=1$
$S_{16}=\frac{16}{2}[2\times1+(16-1)\times\frac{1}{2}]=8\times[2+\frac{15}{2}]=8\times\frac{19}{2}=76$
When $d=-\frac{1}{2}$: $a=5$
$S_{16}=\frac{16}{2}[2\times5+(16-1)\times(-\frac{1}{2})]=8\times[10-\frac{15}{2}]=8\times\frac{5}{2}=20$
The sum of the first 16 terms is 76 or 20.
3. A ladder has rungs 25 cm apart. The rungs decrease uniformly from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are $2\frac{1}{2}\ \text{m}$ apart, what is the length of wood required for the rungs?
Number of rungs $= \frac{250\ \text{cm}}{25\ \text{cm}}+1=11$
The rung lengths form an A.P. with $a_1=45,\ a_{11}=25$.
$45+10d=25 \Rightarrow d=-2$
$S_{11}=\frac{11}{2}[45+25]=\frac{11}{2}\times70=385\ \text{cm}$
The total length of wood required is 385 cm.
The rung lengths form an A.P. with $a_1=45,\ a_{11}=25$.
$45+10d=25 \Rightarrow d=-2$
$S_{11}=\frac{11}{2}[45+25]=\frac{11}{2}\times70=385\ \text{cm}$
The total length of wood required is 385 cm.
4. 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 house number $x$ equals the sum of the numbers following it. Find $x$.
We need $S_{x-1}=S_{49}-S_x$.
$\frac{x-1}{2}[2+(x-2)\times1]=\frac{49}{2}[2+48]-\frac{x}{2}[2+(x-1)\times1]$
$\Rightarrow (x-1)(x)=49\times50-x(x+1)$
$\Rightarrow x^2-x=2450-x^2-x$
$\Rightarrow 2x^2=2450 \Rightarrow x^2=1225 \Rightarrow x=35$
The required value is $\mathbf{x=35}$.
$\frac{x-1}{2}[2+(x-2)\times1]=\frac{49}{2}[2+48]-\frac{x}{2}[2+(x-1)\times1]$
$\Rightarrow (x-1)(x)=49\times50-x(x+1)$
$\Rightarrow x^2-x=2450-x^2-x$
$\Rightarrow 2x^2=2450 \Rightarrow x^2=1225 \Rightarrow x=35$
The required value is $\mathbf{x=35}$.
5. A small terrace at a football ground comprises 15 steps, each 50 m long, with a rise of $\frac{1}{4}\ \text{m}$ and a tread of $\frac{1}{2}\ \text{m}$. Calculate the total volume of concrete required to build the terrace.
Volume for step 1: $\frac{1}{4}\times\frac{1}{2}\times50\ \text{m}^3$
Volume for step 2: $\frac{2}{4}\times\frac{1}{2}\times50\ \text{m}^3$
Volume for step 3: $\frac{3}{4}\times\frac{1}{2}\times50\ \text{m}^3$
These form an A.P. with $a=\frac{50}{8}$ and $d=\frac{50}{8}$.
$S_{15}=\frac{15}{2}\left[\frac{100}{8}+14\times\frac{50}{8}\right]=\frac{15}{2}\times\frac{100+700}{8}=\frac{15}{2}\times100=750\ \text{m}^3$
The total volume of concrete required is $750\ \text{m}^3$.
Volume for step 2: $\frac{2}{4}\times\frac{1}{2}\times50\ \text{m}^3$
Volume for step 3: $\frac{3}{4}\times\frac{1}{2}\times50\ \text{m}^3$
These form an A.P. with $a=\frac{50}{8}$ and $d=\frac{50}{8}$.
$S_{15}=\frac{15}{2}\left[\frac{100}{8}+14\times\frac{50}{8}\right]=\frac{15}{2}\times\frac{100+700}{8}=\frac{15}{2}\times100=750\ \text{m}^3$
The total volume of concrete required is $750\ \text{m}^3$.
