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NCERT Solutions For Class 10 Maths Chapter 5 Arithmetic Progressions
Imagine a little boy saving Rs. 10 daily and putting it in this piggy bank. After a while, this type of savings forms a pattern and increases by a fixed amount every day. This pattern is an arithmetic progression (AP)—a fundamental algebraic concept that underpins various advanced topics. An arithmetic progression is a sequence of numbers in which the difference between each number with respect to the two consecutive numbers is the same. This difference is a key element in identifying the arithmetic progression and its technical properties. Arithmetic progressions have numerous real-life applications, such as finance, physics, computer science, and many other topics. These NCERT Solutions for Class 10 Maths provide an easy-to-follow, step-by-step answer, so that students have the opportunity to understand the exercise problems presented to them in the NCERT Class 10 Maths Book. Many toppers rely on NCERT Solutions since they are designed as per the latest syllabus.
This Story also Contains
- NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions PDF Free Download
- NCERT Solutions for Class 10 Maths Chapter 5: Exercise Questions
- Arithmetic Progressions Class 10 Chapter 5: Topics
- Class 10 Maths Chapter 5 Arithmetic Progressions Notes
- NCERT Solutions for Class 10 Maths: Chapter Wise
Every number in an arithmetic progression has a purpose, just like every effort counts in success. These NCERT solutions for Class 10 have been prepared by Careers360 Subject Matter Experts according to the up-to-date NCERT syllabus, as the development of Arithmetic Progressions involves the development of significant understandings. In addition, after completing the exercises, students can refer to the Arithmetic Progression Notes for quick revision purposes, as well as the NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression to practice for more clarity. Refer to this NCERT webpage for NCERT Syllabus, notes, and PDF resources that may help you keep it up to date.
NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions PDF Free Download
The NCERT Solutions for Class 10 Maths Chapter 5 have been prepared by Careers360 experts to make learning simpler and to help you score better in exams. You can also download the solutions in PDF format.
NCERT Solutions for Class 10 Maths Chapter 5: Exercise Questions
Below are the NCERT class 10 maths chapter 5 solutions for exercise questions.
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Answer: It is given that Fare for $1^{st} \ km$ = Rs. 15 And after that $\small Rs\hspace{1mm}8$ for each additional $\small km$ Now, Fare for $2^{nd} \ km$ = Fare of first km + Additional fare for 1 km = Rs. 15 + 8 = Rs 23 Fare for $3^{rd} \ km$ = Fare of first km + Fare of additional second km + Fare of additional third km = Rs. 23 + 8= Rs 31 Fare of n km = $15 + 8 \times (n - 1)$ ( We multiplied by n - 1 because the first km was fixed and for the rest, we are adding additional fare. In this, each subsequent term is obtained by adding a fixed number (8) to the previous term.) Answer: It is given that Vacum pump removes $\small \frac{1}{4}$ of the air remaining in the cylinder at a time Let us take an initial quantity of air = 1 Now, the quantity of air removed in the first step = 1/4 Remaining quantity after 1st step $= 1-\frac{1}{4}= \frac{3}{4}$ Similarly, Quantity removed after 2nd step = Quantity removed in first step $\times$ Remaining quantity after 1st step $=\frac{3}{4}\times \frac{1}{4}= \frac{3}{16}$ Remaining quantity after 2nd step would be = Remaining quantity after 1st step - Quantity removed after 2nd step $=\frac{3}{4}- \frac{3}{16}= \frac{12-3}{16}= \frac{9}{16}$ After the second step, the difference between the second and first and first and initial step is not the same, hence The common difference (d) is not the same after every step Therefore, it is not an AP. Answer: It is given that Cost of digging of 1st meter = Rs 150 And Rises by $\small Rs\hspace{1mm}50$ for each subsequent meter Therefore, Cost of digging of first 2 meters = cost of digging of first meter + cost of digging the additional meter Cost of digging of first 2 meters = 150 + 50 = Rs 200 Similarly, Cost of digging of first 3 meters = cost of digging of first 2 meters + cost of digging of additional meter Cost of digging of first 3 meters = 200 + 50 = Rs 250 We can clearly see that 150, 200,250, ... is in AP with each subsequent term obtained by adding a fixed number (50) to the previous term. Therefore, it is an AP with first term (a) = 150 and common difference (d) = 50 Answer: Amount in the beginning = Rs. 10000 Interest at the end of 1st year at the rate of $\small 8\hspace{1mm}\%$ is $\small 8\hspace{1mm}\%$ of 10000 = $\frac{8\times 10000}{100}= 800$ Therefore, the amount at the end of 1st year will be = 10000 + 800 = 10800 Now, Interest at the end of 2nd year at rate of $\small 8\hspace{1mm}\%$ is $\small 8\hspace{1mm}\%$ of 10800 = $\frac{8\times 10800}{100}= 864$ Therefore, the amount at the end of 2nd year = 10800 + 864 = 11664 Since each subsequent term is not obtained by adding a unique number to the previous term; hence, it is not an AP Answer: It is given that $\small a=10,d=10$ Now, $a_1= a =10$ $a_2= a_1 + d= 10 + 10 = 20$ $a_3= a_2 + d= 20 + 10 = 30$ $a_4= a_3 + d= 30 + 10 = 40$ Therefore, the first four terms of the given series are 10, 20, 30, 40. Q2 (ii): Write first four terms of the AP when the first term a and the common difference d are given as follows: $\small a=-2,d=0$ It is given that $\small a=-2,d=0$ Now, $a_1= a = -2$ $a_2= a_1 + d= -2 + 0 = -2$ $a_3= a_2 + d= -2 + 0 = -2$ $a_4= a_3 + d= -2 + 0 = -2$ Therefore, the first four terms of the given series are -2, -2, -2, -2. Answer: It is given that $\small a=4,d=-3$ Now, $a_1= a =4$ $a_2= a_1 + d= 4 - 3 = 1$ $a_3= a_2 + d= 1 - 3 = -2$ $a_4= a_3 + d= -2- 3 = -5$ Therefore, the first four terms of the given series are 4, 1, -2, -5 Answer: It is given that $\small a=-1,d=\frac{1}{2}$ Now, $a_1= a =-1$ $a_2= a_1 + d= -1 + \frac{1}{2} = -\frac{1}{2}$ $a_3= a_2 + d= -\frac{1}{2} + \frac{1}{2} = 0$ $a_4= a_3 + d= 0+\frac{1}{2}= \frac{1}{2}$ Therefore, the first four terms of the given series are $-1,-\frac{1}{2},0, \frac{1}{2}$ Answer: It is given that $\small a=-1.25,d=-0.25$ Now, $a_1= a =-1.25$ $a_2= a_1 + d= -1.25 -0.25= -1.50$ $a_3= a_2 + d= -1.50-0.25=-1.75$ $a_4= a_3 + d= -1.75-0.25=-2$ Therefore, the first four terms of the given series are -1.25, -1.50, -1.75, -2 Q3 (i): For the following APs, write the first term and the common difference $\small 3,1,-1,-3,...$ Answer: Given AP series is $\small 3,1,-1,-3,...$ Now, the first term of this AP series is 3 Therefore, First term of AP series (a) = 3 Now, $a_1=3 \ \ and \ \ a_2 = 1$ And common difference (d) = $a_2-a_1 = 1-3 = -2$ Therefore, the first term and the common difference are 3 and -2, respectively Q3 (ii): For the following APs, write the first term and the common difference: $\small -5,-1,3,7,...$ Answer: Given AP series is $\small -5,-1,3,7,...$ Now, the first term of this AP series is -5 Therefore, First-term of AP series (a) = -5 Now, $a_1=-5 \ \ and \ \ a_2 = -1$ And common difference (d) = $a_2-a_1 = -1-(-5) = 4$ Therefore, the first term and the common difference are -5 and 4, respectively Answer: Given AP series is $\small \frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3},...$ Now, the first term of this AP series is $\frac{1}{3}$ Therefore, The first term of AP series (a) = $\frac{1}{3}$ Now, $a_1=\frac{1}{3} \ \ and \ \ a_2 = \frac{5}{3}$ And common difference (d) = $a_2-a_1 = \frac{5}{3}-\frac{1}{3} = \frac{5-1}{3} =\frac{4}{3}$ Therefore, the first term and the common difference is $\frac{1}{3}$ and $\frac{4}{3}$ respectively Q3 (iv): For the following APs, write the first term and the common difference: $\small 0.6,1.7,2.8,3.9,...$ Answer: Given the AP series is $\small 0.6,1.7,2.8,3.9,...$ Now, the first term of this AP series is 0.6 Therefore, First term of AP series (a) = 0.6 Now, $a_1=0.6 \ \ and \ \ a_2 = 1.7$ And common difference (d) = $a_2-a_1 = 1.7-0.6 = 1.1$ Therefore, the first term and the common difference are 0.6 and 1.1, respectively. Answer: Given series is $\small 2,4,8,12,...$ Now, The first term of this series is = 2 Now, $a_1 = 2 \ \ and \ \ a_2 = 4 \ \ and \ \ a_3 = 8$ $a_2-a_1 = 4-2 = 2$ $a_3-a_2 = 8-4 = 4$ We can see that the difference between the terms is not equal Hence, the given series is not an AP. Answer: Given series is $\small 2,\frac{5}{2},3,\frac{7}{2},...$ Now, The first term of this series is = 2 Now, $a_1 = 2 \ \ and \ \ a_2 = \frac{5}{2} \ \ and \ \ a_3 = 3 \ \ and \ \ a_4 = \frac{7}{2}$ $a_2-a_1 = \frac{5}{2}-2 = \frac{5-4}{2}=\frac{1}{2}$ $a_3-a_2 = 3-\frac{5}{2} = \frac{6-5}{2} = \frac{1}{2}$ $a_4-a_3=\frac{7}{2}-3=\frac{7-6}{2} =\frac{1}{2}$ We can clearly see that the difference between terms are equal and equal to $\frac{1}{2}$ Hence, the given series is in AP. Now, the next three terms are $a_5=a_4+d = \frac{7}{2}+\frac{1}{2} = \frac{8}{2}=4$ $a_6=a_5+d = 4+\frac{1}{2} = \frac{8+1}{2}=\frac{9}{2}$ $a_7=a_6+d =\frac{9}{2} +\frac{1}{2} = \frac{10}{2}=5$ Therefore, the next three terms of the given series are $4,\frac{9}{2} ,5$ Answer: Given series is $\small -1.2,-3.2,-5.2,-7.2,...$ Now, The first term of this series is = -1.2 Now, $a_1 = -1.2 \ \ and \ \ a_2 = -3.2 \ \ and \ \ a_3 = -5.2 \ \ and \ \ a_4 = -7.2$ $a_2-a_1 = -3.2-(-1.2) =-3.2+1.2=-2$ $a_3-a_2 = -5.2-(-3.2) =-5.2+3.2 = -2$ $a_4-a_3=-7.2-(-5.2)=-7.2+5.2=-2$ We can clearly see that the difference between terms are equal and equal to -2 Hence, the given series is in AP. Now, the next three terms are $a_5=a_4+d = -7.2-2 =-9.2$ $a_6=a_5+d = -9.2-2 =-11.2$ $a_7=a_6+d = -11.2-2 =-13.2$ Therefore, the next three terms of the given series are -9.2, -11.2, -13.2 Answer: Given series is $\small -10,-6,-2,2,...$ Now, The first term of this series is = -10 Now, $a_1 = -10 \ \ and \ \ a_2 = -6 \ \ and \ \ a_3 = -2 \ \ and \ \ a_4 = 2$ $a_2-a_1 = -6-(-10) =-6+10=4$ $a_3-a_2 = -2-(-6) =-2+6 = 4$ $a_4-a_3=2-(-2)=2+2=4$ We can clearly see that the difference between the terms are equal and equal to 4 Hence, the given series is in AP. Now, the next three terms are $a_5=a_4+d = 2+4 =6$ $a_6=a_5+d = 6+4=10$ $a_7=a_6+d = 10+4=14$ Therefore, the next three terms of the given series are 6,10,14 Answer: Given series is $\small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...$ Now, The first term of this series is = 3 Now, $a_1 = 3 \ \ and \ \ a_2 = 3+\sqrt2 \ \ and \ \ a_3 = 3+2\sqrt2 \ \ and \ \ a_4 = 3+3\sqrt2$ $a_2-a_1 = 3+\sqrt2-3= \sqrt2$ $a_3-a_2 = 3+2\sqrt2-3-\sqrt2 = \sqrt2$ $a_4-a_3 = 3+3\sqrt2-3-2\sqrt2 = \sqrt2$ We can clearly see that the difference between terms are equal and equal to $\sqrt2$ Hence, given series is in AP Now, the next three terms are $a_5=a_4+d = 3+3\sqrt2+\sqrt2=3+4\sqrt2$ $a_6=a_5+d = 3+4\sqrt2+\sqrt2=3+5\sqrt2$ $a_7=a_6+d = 3+5\sqrt2+\sqrt2=3+6\sqrt2$ Therefore, the next three terms of given series are $3+4\sqrt2, 3+5\sqrt2,3+6\sqrt2$ Answer: Given series is $\small 0.2, 0.22, 0.222, 0.2222,...$ Now, The first term to this series is = 0.2 Now, $a_1 = 0.2 \ \ and \ \ a_2 = 0.22 \ \ and \ \ a_3 = 0.222 \ \ and \ \ a_4 = 0.2222$ $a_2-a_1 = 0.22-0.2=0.02$ $a_3-a_2 = 0.222-0.22=0.002$ We can clearly see that the difference between the terms is not equal. Hence, the given series is not an AP Answer: Given series is $\small 0,-4,-8,-12,...$ Now, the first term to this series is = 0 Now, $a_1 = 0 \ \ and \ \ a_2 = -4 \ \ and \ \ a_3 = -8 \ \ and \ \ a_4 = -12$ $a_2-a_1 = -4-0 =-4$ $a_3-a_2 = -8-(-4) =-8+4 = -4$ $a_4-a_3=-12-(-8)=-12+8=-4$ We can clearly see that the difference between terms are equal and equal to -4 Hence, given series is in AP Now, the next three terms are $a_5=a_4+d = -12-4 =-16$ $a_6=a_5+d = -16-4=-20$ $a_7=a_6+d = -20-4=-24$ Therefore, the next three terms of the given series are -16,-20,-24 Q4 (viii): Which of the following are APs? If they form an AP, find the common difference d and write three more terms. $\small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...$ Answer: Given series is $\small -\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},...$ Now, The first term to this series is = $-\frac{1}{2}$ Now, $a_1 = -\frac{1}{2} \ \ and \ \ a_2 = -\frac{1}{2} \ \ and \ \ a_3 = -\frac{1}{2} \ \ and \ \ a_4 = -\frac{1}{2}$ $a_2-a_1 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$ $a_3-a_2 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$ $a_4-a_3 = -\frac{1}{2}-\left ( -\frac{1}{2} \right ) = -\frac{1}{2}+\frac{1}{2}=0$ We can clearly see that the difference between terms are equal and equal to 0 Hence, given series is in AP Now, the next three terms are $a_5=a_4+d = -\frac{1}{2}+0=-\frac{1}{2}$ $a_6=a_5+d = -\frac{1}{2}+0=-\frac{1}{2}$ $a_7=a_6+d = -\frac{1}{2}+0=-\frac{1}{2}$ Therefore, the next three terms of given series are $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}$ Answer: Given series is $\small 1, 3, 9, 27,...$ Now, The first term to this series is = 1 Now, $a_1 = 1 \ \ and \ \ a_2 = 3 \ \ and \ \ a_3 = 9 \ \ and \ \ a_4 = 27$ $a_2-a_1 = 3-1=2$ $a_3-a_2 =9-3=6$ We can clearly see that the difference between terms are not equal. Hence, given series is not an AP. Answer: Given series is $\small a,2a,3a,4a,...$ Q4 (xi): Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. $\small a,a^2,a^3,a^4,...$ Given series is Answer: Given series is That is the next three terms are $\sqrt{50},\ \sqrt{72},\ \sqrt{98}$ Answer: Given series is Answer: Given series is Answer: Given series is |
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Answer: (iii) Given $d=-3, n = 18, a_{18} = -5$ (A) $\small 97$ (B) $\small 77$ (C) $\small -77$ (D) $\small -87$ Answer: (A) $\small 28$ (B) $\small 22$ (C) $\small -38$ (D) $\small -48\frac{1}{2}$ Answer: Answer: Answer: Answer: Q3 (iv): In the following APs, find the missing terms in the boxes: $\small -4,\: \fbox { },\: \fbox { },\: \fbox { },\: \fbox { },\: 6$ Given AP series is $\small -4,\: \fbox { },\: \fbox { },\: \fbox { },\: \fbox { },\: 6$ Answer: Q4: Which term of the AP : $\small 3,8,13,18,...,$ is $\small 78$ ? Q5 (i): Find the number of terms in each of the following APs: $\small 7,13,19,...,205$ Answer: Q5 (ii): Find the number of terms in each of the following APs: $\small 18,15\frac{1}{2},13,...,-47$ Answer: Q6: Check whether $\small -150$ is a term of the AP : $\small 11,8,5,2...$ Answer: Answer: Q8: An AP consists of $\small 50$ terms of which $\small 3$ rd term is $\small 12$ and the last term is $\small 106$ . Find the $\small 29$ th term. Q9: If the $\small 3$ rd and the $\small 9$ th terms of an AP are $\small 4$ and $\small -8$ respectively, which term of this AP is zero? Q10: The $\small 17$ th term of an AP exceeds its $\small 10$ th term by $\small 7$ . Find the common difference. Q11: Which term of the AP : $\small 3,15,27,39,...$ will be $\small 132$ more than its $\small 54$ th term? Q12: Two APs have the same common difference. The difference between their $\small 100$ th terms is $\small 100$ , what is the difference between their $\small 1000$ th terms? Q13: How many three-digit numbers are divisible by $\small 7$ ? Q14: How many multiples of $\small 4$ lie between $\small 10$ and $\small 250$ ? Answer: Q16: Determine the AP whose third term is $\small 16$ and the $\small 7$ th term exceeds the $\small 5$ th term by $\small 12$ . Q17: Find the $\small 20$ th term from the last term of the AP : $\small 3,8,13,...,253$ . Answer: Q18: The sum of the $\small 4$ th and $\small 8$ th terms of an AP is $\small 24$ and the sum of the $\small 6$ th and $\small 10$ th terms is $\small 44$. Find the first three terms of the AP. Answer: Answer: |
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Q1 (i): Find the sum of the following APs: $\small 2,7,12,...,$ to $\small 10$ terms. Answer: Given series $\small 2,7,12,...,$ to $\small 10$ terms Here, $a = 2 \ and \ n = 10$ and $d = a_2-a_1=7-2=5$ Now, we know that $S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S = \frac{10}{2}\left \{ 2\times 2 +(10-1)5\right \}$ $\Rightarrow S = 5\left \{ 4 +45\right \}$ $\Rightarrow S = 5\left \{ 49\right \}$ $\Rightarrow S =245$ Therefore, the sum of AP $\small 2,7,12,...,$ to $\small 10$ terms is 245 Q1 (ii); Find the sum of the following APs: $\small -37,-33,-29,...,$ to $\small 12$ terms. Answer: Given series $\small -37,-33,-29,...,$ to $\small 12$ terms. Here, $a = -37 \ and \ n = 12$ and $d = a_2-a_1=-33-(-37)=4$ Now, we know that $S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S = \frac{12}{2}\left \{ 2\times (-37) +(12-1)4\right \}$ $\Rightarrow S = 6\left \{ -74 +44\right \}$ $\Rightarrow S = 5\left \{ -30\right \}$ $\Rightarrow S =-180$ Therefore, the sum of AP $\small -37,-33,-29,...,$ to $\small 12$ terms. is -180 Q1 (iii): Find the sum of the following APs: $\small 0.6,1.7,2.8,...,$ to $\small 100$ terms. Answer: Given series $\small 0.6,1.7,2.8,...,$ to $\small 100$ terms.. Here, $a = 0.6 \ and \ n = 100$ and $d = a_2-a_1=1.7-0.6=1.1$ Now, we know that $S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S = \frac{100}{2}\left \{ 2\times (0.6) +(100-1)(1.1)\right \}$ $\Rightarrow S = 50\left \{ 1.2 +108.9\right \}$ $\Rightarrow S = 50\left \{ 110.1\right \}$ $\Rightarrow S =5505$ Therefore, the sum of AP $\small 0.6,1.7,2.8,...,$ to $\small 100$ terms. is 5505 Answer: Given series $\small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,$ to $\small 11$ terms. Here, $a = \frac{1}{15} \ and \ n = 11$ and $d = a_2-a_1=\frac{1}{12}-\frac{1}{15}= \frac{5-4}{60}= \frac{1}{60}$ Now, we know that $S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S = \frac{11}{2}\left \{ 2\times \frac{1}{15} +(11-1)(\frac{1}{60})\right \}$ $\Rightarrow S = \frac{11}{2}\left \{ \frac{2}{15} +\frac{1}{6}\right \}$ $\Rightarrow S = \frac{11}{2}\left \{ \frac{9}{30}\right \}$ $\Rightarrow S =\frac{99}{60}= \frac{33}{20}$ Therefore, the sum of AP $\small \frac{1}{15},\frac{1}{12},\frac{1}{10},...,$ to $\small 11$ terms. is $\frac{33}{20}$ Q2 (i): Find the sums given below: $\small 7+10\frac{1}{2}+14+...+84$ Answer: Given series $\small 7+10\frac{1}{2}+14+...+84$ First we need to find the number of terms Here, $a = 7 \ and \ a_n = 84$ and $d = a_2-a_1=\frac{21}{2}-7= \frac{21-14}{2}= \frac{7}{2}$ Let's suppose there are n terms in the AP Now, we know that $a_n = a+(n-1)d$ $\Rightarrow 84 = 7 + (n-1)\frac{7}{2}$ $\Rightarrow \frac{7n}{2}= 77+\frac{7}{2}$ $\Rightarrow n = 23$ Now, we know that $S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S = \frac{23}{2}\left \{ 2\times7 +(23-1)(\frac{7}{2})\right \}$ $\Rightarrow S = \frac{23}{2}\left \{ 14 +77\right \}$ $\Rightarrow S = \frac{23}{2}\left \{ 91\right \}$ $\Rightarrow S =\frac{2093}{2}=1046\frac{1}{2}$ Therefore, the sum of AP $\small 7+10\frac{1}{2}+14+...+84$ is $1046\frac{1}{2}$ Q2 (ii): Find the sums given below: $\small 34+32+30+...+10$ Answer: Given series $\small 34+32+30+...+10$ First we need to find the number of terms Here, $a = 34 \ and \ a_n = 10$ and $d = a_2-a_1=32-34=-2$ Let's suppose there are n terms in the AP Now, we know that $a_n = a+(n-1)d$ $\Rightarrow 10 = 34 + (n-1)(-2)$ $\Rightarrow -26 = -2n$ $\Rightarrow n = 13$ Now, we know that $S = \frac{n}{2}\left \{ a+a_n \right \}$ $\Rightarrow S =13\times 22 = 286$ Therefore, the sum of AP $\small 34+32+30+...+10$ is 286 Q2 (iii): Find the sums given below: $\small -5+(-8)+(-11)+...+(-230)$ Answer: Given series $\small -5+(-8)+(-11)+...+(-230)$ First, we need to find the number of terms Here, $a = -5 \ and \ a_n = -230$ and $d = a_2-a_1=-8-(-5)= -3$ Let suppose there are n terms in the AP Now, we know that $a_n = a+(n-1)d$ $\Rightarrow -230 = -5 + (n-1)(-3)$ $\Rightarrow -228 = -3n$ $\Rightarrow n = 76$ Now, we know that $S = \frac{n}{2}\left \{ a+a_n \right \}$ $\Rightarrow S = \frac{76}{2}\left \{ (-5-230 )\right \}$ $\Rightarrow S = 38\left \{ -235\right \}$ $\Rightarrow S = -8930$ Therefore, the sum of AP $\small -5+(-8)+(-11)+...+(-230)$ is -8930 Q3 (i): In an AP: given $\small a=5$ , $\small d=3$ , $\small a_n=50$ , find $\small n$ and $\small S_n$ . Answer: Given $a = 5, d = 3 \ and \ a_n = 50$ Let's suppose there are n terms in the AP Now, we know that $a_n = a+(n-1)d$ $\Rightarrow 50 = 5 + (n-1)3$ $\Rightarrow 48 = 3n$ $\Rightarrow n = 16$ Now, we know that $S = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S = \frac{16}{2}\left \{ 2\times(5) +(16-1)(3)\right \}$ $\Rightarrow S = 8\left \{ 10+45\right \}$ $\Rightarrow S = 8\left \{ 55\right \}$ $\Rightarrow S =440$ Therefore, the sum of the given AP is 440 Q3 (ii): In an AP: given $\small a=7$ , $a_{13}=35$ , find $d$ and $S_{13}$ . Answer: Given $a = 7 \ and \ a_{13} = 35$ $a_{13}= a+12d = 35$ $= 12d = 35-7 = 28$ $d = \frac{28}{12}= \frac{7}{3}$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{13} = \frac{13}{2}\left \{ 2\times(7) +(13-1)(\frac{7}{3})\right \}$ $\Rightarrow S_{13} = \frac{13}{2}\left \{14 +28\right \}$ $\Rightarrow S_{13} = \frac{13}{2}\left \{42\right \}$ $\Rightarrow S_{13} = 13 \times 21 = 273$ Therefore, the sum of given AP is 273 Q3 (iii): In an AP: given $a_{12}=37,d=3,$ find $\small a$ and $S_{12}$ . Answer: Given $d = 3 \ and \ a_{12} = 37$ $a_{12}= a+11d = 37$ $= a= 37-11\times 3 = 37-33=4$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{12} = \frac{12}{2}\left \{ 2\times(4) +(12-1)3\right \}$ $\Rightarrow S_{12} = 6\left \{ 8+33\right \}$ $\Rightarrow S_{12} = 6\left \{41\right \}$ $\Rightarrow S_{12} =246$ Therefore, the sum of given AP is 246 Q3 (iv): In an AP: given $a_3=15, S_{10}=125,$ find $\small d$ and $a_{10}$ Answer: Given $a_3=15, S_{10}=125$ $a_{3}= a+2d = 15 \ \ \ \ \ \ \ \ -(i)$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{10} = \frac{10}{2}\left \{ 2\times(a) +(10-1)d\right \}$ $\Rightarrow 125 = 5\left \{ 2a+9d\right \}$ $\Rightarrow 2a+9d = 25 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$ On solving equation (i) and (ii) we will get $a= 17 \ and \ d = -1$ Now, $a_{10} = a+ 9d = 17 + 9(-1)= 17-9 = 8$ Therefore, the value of d and 10th terms is -1 and 8, respectively Q3 (v): In an AP: given $\small d=5, S_9=75$ , find $\small a$ and $\small a_9$ . Answer: Given $\small d=5, S_9=75$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{9} = \frac{9}{2}\left \{ 2\times(a) +(9-1)5\right \}$ $\Rightarrow 75= \frac{9}{2}\left \{ 2a +40\right \}$ $\Rightarrow 150= 18a+360$ $\Rightarrow a = -\frac{210}{18}=-\frac{35}{3}$ Now, $a_{9} = a+ 8d = -\frac{35}{3} + 8(5)= -\frac{35}{3}+40 = \frac{-35+120}{3}= \frac{85}{3}$ Q3 (vi): In an AP: given $\small a=2,d=8,S_n=90,$ find $\small n$ and $\small a_n$ . Answer: Given $\small a=2,d=8,S_n=90,$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow 90 = \frac{n}{2}\left \{ 2\times(2) +(n-1)8\right \}$ $\Rightarrow 180 = n\left \{ 4+8n-8\right \}$ $\Rightarrow 8n^2-4n-180=0$ $\Rightarrow 4(2n^2-n-45)=0$ $\Rightarrow 2n^2-n-45=0$ $\Rightarrow 2n^2-10n+9n-45=0$ $\Rightarrow (n-5)(2n+9)=0$ $\Rightarrow n = 5 \ \ and \ \ n = - \frac{9}{2}$ We know that 'n' can not be negative so the only the value of n is 5 Now, $a_{5} = a+ 4d = 2+4\times 8 = 2+32 = 34$ Therefore, value of n and nth term is 5 and 34 respectively Q3 (vii): In an AP: given $\small a=8,a_n=62,S_n=210,$ find $\small n$ and $\small d$. Answer: Given $\small a=8,a_n=62,S_n=210,$ Now, we know that $a_n = a+(n-1)d$ $62 = 8+(n-1)d$ $(n-1)d= 54 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow 210 = \frac{n}{2}\left \{ 2\times(8) +(n-1)d\right \}$ $\Rightarrow 420 = n\left \{ 16+54 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$ $\Rightarrow 420 = n\left \{ 70 \right \}$ $\Rightarrow n = 6$ Now, put this value in (i) we will get $d = \frac{54}{5}$ Therefore, value of n and d are $6 \ and \ \frac{54}{5}$ respectively Q3 (viii): In an AP: given $\small a_n=4,d=2,S_n=-14,$ find $\small n$ and $\small a$ . Answer: Given $\small a_n=4,d=2,S_n=-14,$ Now, we know that $a_n = a+(n-1)d$ $4 = a+(n-1)2$ $a+2n = 6\Rightarrow a = 6-2n \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow -14 = \frac{n}{2}\left \{ 2\times(a) +(n-1)2\right \}$ $\Rightarrow -28 = n\left \{ 2(6-2n)+2n-2 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$ $\Rightarrow -28 = n\left \{ 10-2n \right \}$ $\Rightarrow 2n^2-10n-28=0$ $\Rightarrow 2(n^2-5n-14)=0$ $\Rightarrow n^2-7n+2n-14=0$ $\Rightarrow(n+2)(n-7)=0$ $\Rightarrow n = -2 \ \ and \ \ n = 7$ Value of n cannot be negative so the only the value of n is 7 Now, put this value in (i) we will get a = -8 Therefore, the value of n and a are 7 and -8 respectively Q3 (ix): In an AP: given $\small a=3,n=8,S=192,$ find $\small d$ . Answer: Given $\small a=3,n=8,S=192,$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow 192 = \frac{8}{2}\left \{ 2\times(3) +(8-1)d\right \}$ $\Rightarrow 192=4\left \{6 +7d\right \}$ $\Rightarrow 7d = 48-6$ $\Rightarrow d = \frac{42}{7} = 6$ Therefore, the value of d is 6 Answer: Given $\small l=28,S=144 \ and \ n = 9$ Now, we know that $l = a_n = a+(n-1)d$ $28 = a_n = a+(n-1)d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow 144 = \frac{9}{2}\left \{ a + a +(n-1)d\right \}$ $\Rightarrow 288 =9\left \{ a+ 28\right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(using \ (i))$ $\Rightarrow a+28= 32$ $\Rightarrow a=4$ Therefore, the value of a is 4 Q4: How many terms of the AP: $\small 9,17,25,...$ must be taken to give a sum of $\small 636$ ? Answer: Given series $\small 9,17,25,...$ Here, $a =9 \ and \ d = 8$ And $S_n = 636$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ After putting values we get: $\Rightarrow 636 = \frac{n}{2}\left \{ 18+(n-1)8 \right \}$ $\Rightarrow 1272 = n\left \{ 10+8n \right \}$ $\Rightarrow 8n^2+10n-1272=0$ $\Rightarrow 2(4n^2+5n-636)=0$ $\Rightarrow 4n^2+53n-48n-636=0$ $\Rightarrow (n-12)(4n+53)=0$ $\Rightarrow n = 12 \ \ and \ \ n = - \frac{53}{4}$ The value of n can not be negative, so the only value of n is 12 Therefore, the sum of 12 terms of AP $\ small9,17,25,...$ must be taken to give a sum of $\small 636$. Answer: Given $\small a=5,a_n=45,S_n=400,$ Now, we know that $a_n = a+(n-1)d$ $45 = 5+(n-1)d$ $(n-1)d= 40 \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow 400 = \frac{n}{2}\left \{ 2\times(5) +(n-1)d\right \}$ $\Rightarrow 800 = n\left \{ 10+40 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$ $\Rightarrow 800 = n\left \{ 50 \right \}$ $\Rightarrow n = 16$ Now, put this value in (i), we will get $d = \frac{40}{15}= \frac{8}{3}$ Therefore, value of n and d are $16 \ and \ \frac{8}{3}$ respectively Answer: Given $\small a=17,l=350,d=9,$ Now, we know that $a_n = a+(n-1)d$ $350 = 17+(n-1)9$ $(n-1)9 = 333$ $(n-1)=37$ $n = 38$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{38}= \frac{38}{2}\left \{ 2\times(17) +(38-1)9\right \}$ $\Rightarrow S_{38}= 19\left \{ 34 +333\right \}$ $\Rightarrow S_{38}= 19\left \{367\right \}$ $\Rightarrow S_{38}= 6973$ Therefore, there are 38 terms and their sum is 6973. Answer: Given $\small a_{22}=149,d=7,n = 22$ Now, we know that $a_{22} = a+21d$ $149 = a+21\times 7$ $a = 149 - 147 = 2$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{22}= \frac{22}{2}\left \{ 2\times(2) +(22-1)7\right \}$ $\Rightarrow S_{22}= 11\left \{ 4 +147\right \}$ $\Rightarrow S_{22}= 11\left \{ 151\right \}$ $\Rightarrow S_{22}= 1661$ Therefore, there are 22 terms and their sum is 1661. Answer: Given $\small a_{2}=14,a_3=18,n = 51$ and $d= a_3-a_2= 18-14=4$ Now, $a_2 = a+d$ $a= 14-4 = 10$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{51}= \frac{51}{2}\left \{ 2\times(10) +(51-1)4\right \}$ $\Rightarrow S_{51}= \frac{51}{2}\left \{ 20 +200\right \}$ $\Rightarrow S_{51}= \frac{51}{2}\left \{ 220\right \}$ $\Rightarrow S_{51}= 51 \times 110$ $\Rightarrow S_{51}=5304$ Therefore, there are 51 terms and their sum is 5610. Answer: Given $S_7 = 49 \ and \ S_{17}= 289$ Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ After putting values we get: $\Rightarrow S_{7}= \frac{7}{2}\left \{ 2\times(a) +(7-1)d\right \}$ $\Rightarrow 98= 7\left \{ 2a +6d\right \}$ $\Rightarrow a +3d = 7 \ \ \ \ \ \ \ -(i)$ Similarly, $\Rightarrow S_{17}= \frac{17}{2}\left \{ 2\times(a) +(17-1)d\right \}$ $\Rightarrow 578= 17\left \{ 2a +16d\right \}$ $\Rightarrow a +8d = 17 \ \ \ \ \ \ \ -(ii)$ On solving equations (i) and (ii), we will get a = 1 and d = 2 Now, the sum of the first n terms is $S_n = \frac{n}{2}\left \{ 2\times 1 +(n-1)2 \right \}$ $S_n = \frac{n}{2}\left \{ 2 +2n-2 \right \}$ $S_n = n^2$ Therefore, the sum of n terms is $n^2$. Answer: Given $\small a_n=3+4n$ We will check the values of $a_n$ for different values of n $a_1 = 3+4(1) =3+4= 7$ $a_2 = 3+4(2) =3+8= 11$ $a_3 = 3+4(3) =3+12= 15$ and so on. From the above, we can see that this is an AP with the first term(a) equal to 7 and a common difference (d) equal to 4 Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times(7) +(15-1)4\right \}$ $\Rightarrow S_{15}= \frac{15}{2}\left \{ 14 +56\right \}$ $\Rightarrow S_{15}= \frac{15}{2}\left \{ 70\right \}$ $\Rightarrow S_{15}= 15 \times 35$ $\Rightarrow S_{15}= 525$ Therefore, the sum of 15 terms is 525. Answer: Given $\small a_n=9-5n$ We will check the values of $a_n$ for different values of n $a_1 = 9-5(1) =9-5= 4$ $a_2 = 9-5(2) =9-10= -1$ $a_3 = 9-5(3) =9-15= -6$ and so on. From the above, we can see that this is an AP with the first term (a) equal to 4 and common difference (d) equal to -5 Now, we know that $S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times(4) +(15-1)(-5)\right \}$ $\Rightarrow S_{15}= \frac{15}{2}\left \{ 8 -70\right \}$ $\Rightarrow S_{15}= \frac{15}{2}\left \{ -62\right \}$ $\Rightarrow S_{15}= 15 \times (-31)$ $\Rightarrow S_{15}= -465$ Therefore, the sum of 15 terms is -465. Answer: Given that the sum of the first $\small n$ terms of an AP is $\small 4n-n^2$ Now, $\Rightarrow S_n = 4n-n^2$ Now, first term is $\Rightarrow S_1 = 4(1)-1^2=4-1=3$ Therefore, first term is 3 $\Rightarrow S_2 = 4(2)-2^2=8-4=4$ Therefore, sum of first two terms is 4 Now, we know that $\Rightarrow S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_2 = \frac{2}{2}\left \{ 2\times 3+(2-1)d \right \}$ $\Rightarrow 4 = \left \{6+d \right \}$ $\Rightarrow d = -2$ Now, $a_2= a+d = 3+(-2 )= 1$ Similarly, $a_3= a+2d = 3+2(-2 )= 3-4=-1$ $a_{10}= a+9d = 3+9(-2 )= 3-18=-15$ $a_{n}= a+(n-1)d = 3+(n-1)(-2 )= 5-2n$ Q12: Find the sum of the first $\small 40$ positive integers divisible by $\small 6$ . Answer: Positive integers divisible by 6 are 6,12,18,... Thus, this is an AP with $here, \ a = 6 \ and \ d = 6$ Now, we know that $S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{40}= \frac{40}{2}\left \{ 2\times 6+(40-1)6 \right \}$ $\Rightarrow S_{40}= 20\left \{12+234 \right \}$ $\Rightarrow S_{40}= 20\left \{246 \right \}$ $\Rightarrow S_{40}= 4920$ Therefore, sum of the first $\small 40$ positive integers divisible by $\small 6$ is 4920. Q13: Find the sum of the first $\small 15$ multiples of $\small 8$ . Answer: The first 15 multiples of 8 are 8,16,24,... Therefore, this is an AP with $here, \ a = 8 \ and \ d = 8$ Now, we know that $S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{15}= \frac{15}{2}\left \{ 2\times 8+(15-1)8 \right \}$ $\Rightarrow S_{15}= \frac{15}{2}\left \{ 16+112 \right \}$ $\Rightarrow S_{15}= \frac{15}{2}\left \{ 128 \right \}$ $\Rightarrow S_{15}= 15 \times 64 = 960$ Therefore, the sum of the first 15 multiples of 8 is 960. Q14: Find the sum of the odd numbers between $\small 0$ and $\small 50$ . Answer: The odd numbers between 0 and 50 are 1,3,5,...49 This is an AP with $here, \ a = 1 \ and \ d = 2$ There are a total of 25 odd numbers between 0 and 50 Now, we know that $S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}$ $\Rightarrow S_{25}= \frac{25}{2}\left \{ 2\times 1+(25-1)2 \right \}$ $\Rightarrow S_{25}= \frac{25}{2}\left \{ 2+48 \right \}$ $\Rightarrow S_{25}= \frac{25}{2}\times 50$ $\Rightarrow S_{25}= 25 \times 25 = 625$ Therefore, sum of the odd numbers between $\small 0$ and $\small 50$ 625 Answer: Given: Penalty for delay of completion beyond a certain date is Rs $\small 200$ for the first day, Rs $\small 250$ for the second day, Rs $\small 300$ for the third day, and a penalty for each succeeding day being Rs $\small 50$ more than for the preceding day We can see that 200,250,300,..... is an AP with $a = 200 \ and \ d = 50$ Now, the penalty for 30 days is given by the expression $S_{30}= \frac{30}{2}\left \{ 2\times 200+(30-1)50 \right \}$ $S_{30}= 15\left ( 400+1450 \right )$ $S_{30}= 15 \times 1850$ $S_{30}= 27750$ Therefore, the penalty for 30 days is 27750 Answer: Given: each price is decreased by 20 rupees, Therefore, d = -20 and there are total 7 prizes so n = 7 and sum of prize money is Rs 700 so $S_7 = 700$ Let a be the prize money given to the 1st student Then, $S_7 = \frac{7}{2}\left \{ 2a+(7-1)(-20) \right \}$ $700 = \frac{7}{2}\left \{ 2a-120 \right \}$ $2a - 120 = 200$ $a = \frac{320}{2}= 160$ Therefore, the prize given to the first student is Rs 160 Now, Let $a_2,a_2,...,a_7$ is the prize money given to the next 6 students Thus, $a_2 = a+d = 160+(-20)=160-20=140$ $a_3 = a+2d = 160+2(-20)=160-40=120$ $a_4 = a+3d = 160+3(-20)=160-60=100$ $a_5 = a+4d = 160+4(-20)=160-80=80$ $a_6 = a+5d = 160+5(-20)=160-100=60$ $a_7 = a+6d = 160+6(-20)=160-120=40$ Therefore, the prize money given to 1 to 7 students is 160, 140, 120, 100, 80, 60, 40. Answer: First, there are 12 classes,s and each class has 3 sections Since each section of class 1 will plant 1 tree, so 3 trees will be planted by 3 sections of class 1. Thus, every class will plant 3 times the number of their class Similarly, No. of trees planted by 3 sections of class 1 = 3 Number of trees planted by 3 sections of class 2 = 6 No. of trees planted by 3 sections of class 3 = 9 Number of trees planted by 3 sections of class 4 = 12 Its clearly an AP with first term (a) = 3 and common difference (d) = 3 and total number of classes (n) = 12 Now, the number of trees planted by 12 classes is given by $S_{12}= \frac{12}{2}\left \{ 2\times 3+(12-1)\times 3 \right \}$ $S_{12}= 6\left ( 6+33 \right )$ $S_{12}= 6 \times 39 = 234$ Therefore, the number of trees planted by 12 classes is 234. |
[ Hint : Length of successive semicircles is $\small l_1,l_2,l_3,l_4,...$ with centres at $\small A,B,A,B,...,$ respectively.]
Answer:
From the above-given figure
Circumference of 1st semicircle $l_1 = \pi r_1 = 0.5\pi$
Similarly,
Circumference of 2nd semicircle $l_2 = \pi r_2 = \pi$
Circumference of 3rd semicircle $l_3 = \pi r_3 = 1.5\pi$
It is clear that this is an AP with $a = 0.5\pi \ and \ d = 0.5\pi$
Now, the sum of the lengths of 13 such semicircles is given by
$S_{13} = \frac{13}{2}\left \{ 2\times 0.5\pi + (13-1)0.5\pi\right \}$
$S_{13} = \frac{13}{2}\left ( \pi+6\pi \right )$
$S_{13} = \frac{13}{2}\times 7\pi$
$S_{13} = \frac{91\pi}{2} = \frac{91}{2}\times \frac{22}{7}=143$
Therefore, sum of length of 13 such semicircles is 143 cm.
Q19: $\small 200$ logs are stacked in the following manner: $\small 20$ logs in the bottom row, $\small 19$ in the next row, $\small 18$ in the row next to it and so on (see Fig. $\small 5.5$ ). In how many rows are the $\small 200$ logs placed and how many logs are in the top row?
Answer:
As the rows are going up, the number of logs is decreasing,
We can clearly see that 20, 19, 18, ..., is an AP and here $a = 20 \ and \ d = -1$
Let's suppose 200 logs are arranged in 'n' rows,
Then,
$S_n = \frac{n}{2}\left \{ 2\times 20 +(n-1)(-1) \right \}$
$200 = \frac{n}{2}\left \{ 41-n \right \}$
$\Rightarrow n^2-41n +400 = 0$
$\Rightarrow n^2-16n-25n +400 = 0$
$\Rightarrow (n-16)(n-25) = 0$
$\Rightarrow n = 16 \ \ and \ \ n = 25$
Now, case (i) n = 25
$a_{25} =a+24d = 20+24\times (-1)= 20-24=-4$
But number of rows can not be in negative numbers
Therefore, we will reject the value n = 25
Case (ii) n = 16
$a_{16} =a+15d = 20+15\times (-1)= 20-15=5$
Therefore, the number of rows in which 200 logs are arranged is equal to 5.
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?
[ Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is $\small 2\times5+2\times (5+3)$ ]
Answer:
Distance travelled by the competitor in picking and dropping 1st potato $= 2 \times 5 = 10 \ m$
Distance travelled by the competitor in picking and dropping 2nd potato $= 2 \times (5+3) =2\times 8 = 16 \ m$
Distance travelled by the competitor in picking and dropping 3rd potato $= 2 \times (5+3+3) =2\times 11 = 22 \ m$ and so on
We can see that it is an AP with first term (a) = 10 and common difference (d) = 6
There are 10 potatoes in the line
Therefore, the total distance travelled by the competitor in picking and dropping potatoes is
$S_{10}= \frac{10}{2}\left \{ 2\times 10+(10-1)6 \right \}$
$S_{10}= 5\left ( 20+54 \right )$
$S_{10}= 5\times 74 = 370$
Therefore, the total distance travelled by the competitor in picking and dropping potatoes is 370 m.
|
Answer: Given AP is $\small 121,117,113,...,$ Here $a = 121 \ and \ d = -4$ Let's suppose the nth term of the AP is the first negative term Then, $a_n = a+ (n-1)d$ If the nth term is negative, then $a_n < 0$ $\Rightarrow 121+(n-1)(-4) < 0$ $\Rightarrow 125<4n$ $\Rightarrow n > \frac{125}{4}=31.25$ Therefore, the first negative term must be the 32nd term Answer: It is given that the sum of the third and seventh terms of an AP is and their product is $\small 8$ $a_3= a+ 2d$ $a_7= a+ 6d$ Now, $a_3+a_7= a+ 2d+a+6d= 6$ $\Rightarrow 2a+8d = 6$ $\Rightarrow a+4d = 3 \Rightarrow a = 3-4d \ \ \ \ \ \ \ \ \ \ \ \ -(i)$ And $a_3.a_7 = (a+2d).(a+6d)=a^2+8ad +12d^2 = 8 \ \ \ \ \ \ \ -(ii)$ Putting the value from equation (i) in equation (ii), we will get $\Rightarrow (3-4d)^2+8(3-4d)d+12d^2= 8$ $\Rightarrow 9+16d^2-24d+24d-32d^2+12d^2=8$ $\Rightarrow 4d^2 = 1$ $\Rightarrow d = \pm \frac{1}{2}$ Now, Case (i) $d = \frac{1}{2}$ $a= 3 - 4 \times \frac{1}{2} = 1$ Then, $S_{16}=\frac{16}{2}\left \{ 2\times 1+(16-1)\frac{1}{2} \right \}$ $S_{16}=76$ Case (ii) $d = -\frac{1}{2}$ $a= 3 - 4 \times \left ( -\frac{1}{2} \right ) = 5$ Then, $S_{16}=\frac{16}{2}\left \{ 2\times 1+(16-1)\left ( -\frac{1}{2} \right ) \right \}$ $S_{16}=20$ Q3: A ladder has rungs $\small 25$ cm apart. (see Fig. $\small 5.7$ ). The rungs decrease uniformly in length from $\small 45$ cm at the bottom to $\small 25$ cm at the top. If the top and the bottom rungs are $\small 2\frac{1}{2}$ m apart, what is the length of the wood required for the rungs? [Hint:Number of rungs = $\frac{250}{25}+1$] |
Answer:
It is given that
The total distance between the top and bottom rung $= 2\frac{1}{2}\ m = 250cm$
Distance between any two rungs = 25 cm
Total number of rungs = $\frac{250}{25}+1= 11$
And it is also given that the bottom-most rung is of 45 cm length and the top-most is of 25 cm length. As it is given that the length of the rungs
decreases uniformly, it will form an AP with $a = 25, a_{11} = 45 \ and \ n = 11$
Now, we know that
$a_{11}= a+ 10d$
$\Rightarrow 45=25+10d$
$\Rightarrow d = 2$
Now, the total length of the wood required for the rungs is equal to
$S_{11} = \frac{11}{2}\left \{ 2\times 25+(11-1)2 \right \}$
$S_{11} = \frac{11}{2}\left \{ 50+20 \right \}$
$S_{11} = \frac{11}{2}\times 70$
$S_{11} =385 \ cm$
Therefore, the total length of the wood required for the rungs is equal to 385 cm
Answer:
It is given that the sum of the numbers of the houses preceding the house numbered $\small x$ is equal to the sum of the numbers of the
houses following it
And 1,2,3,.....,49 form an AP with a = 1 and d = 1
Now, we know that
$S_n= \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
Suppose there exists an n term such that ( n < 49)
Now, according to the given conditions
The sum of the first n - 1 terms of AP = The sum of the terms following the nth term
Sum of the first n - 1 terms of AP = Sum of the whole AP - Sum of the first m terms of AP
i.e.
$S_{n-1}=S_{49}-S_n$
$\frac{n-1}{2}\left \{ 2a+((n-1)-1)d \right \}=\frac{49}{2}\left \{ 2a+(49-1)d \right \}-\frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\frac{n-1}{2}\left \{ 2+(n-2) \right \}=\frac{49}{2}\left \{ 2+48 \right \}-\frac{n}{2}\left \{ 2+(n-1) \right \}$
$\frac{n-1}{2}\left \{ n\right \}=\frac{49}{2}\left \{ 50 \right \}-\frac{n}{2}\left \{ n+1 \right \}$
$\frac{n^2}{2}-\frac{n}{2}=1225-\frac{n^2}{2}-\frac{n}{2}$
$n^2 = 1225$
$n = \pm 35$
Given House numbers are not negative,e so we reject n = -35
Therefore, the sum of no of houses preceding house no 35 is equal to the sum of no of houses following house number 35.
Answer:
It is given that
football ground comprises of $\small 15$ steps each of which is $\small 50$ m long and Each step has a rise of $\small \frac{1}{4}\: m$
and a tread of $\small \frac{1}{2}\: m$
Now,
The volume required to make the first step = $\frac{1}{4}\times \frac{1}{2}\times 50 = 6.25 \ m^3$
Similarly,
The volume required to make 2nd step = $\left ( \frac{1}{4}+\frac{1}{4}\right )\times \frac{1}{2}\times 50=\frac{1}{2}\times \frac{1}{2}\times 50 = 12.5 \ m^3$
And
The volume required to make 3rd step = $\left ( \frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right )\times \frac{1}{2}\times 50=\frac{3}{4}\times \frac{1}{2}\times 50 = 18.75 \ m^3$
And so on
We can see that this is an AP with $a= 6.25 \ and \ d = 6.25$
Now, the total volume of concrete required to build the terrace of 15 such steps is
$S_{15} =\frac{15}{2}\left \{ 2 \times 6.25 +(15-1)6.25 \right \}$
$S_{15} =\frac{15}{2}\left \{ 12.5 +87.5\right \}$
$S_{15} =\frac{15}{2}\times 100$
$S_{15} =15\times 50 = 750$
Therefore, the total volume of concrete required to build the terrace of 15 such steps is $750 \ m^3$.
Also, read,
Arithmetic Progressions Class 10 Chapter 5: Topics
The topics discussed in the NCERT Solutions for class 10, chapter 5, Arithmetic Progressions are:
- Introduction
- Arithmetic Progressions
- nth Term of an AP
- Sum of the First n Terms of an AP
Recommended eBooks
Related eBooks & Sample Papers
Class 10 Maths Chapter 5 Arithmetic Progressions Notes
Arithmetic Progression
Arithmetic Progressions (AP) involve a sequence of terms denoted as $a_1, a_2, a_3,⋯, a_n$ representing a series of integers. A sequence $ a_1, a_2,a_3,⋯, a_n$ is called an arithmetic sequence if $a_{n+1}=a_n+d,n∈N$ where d is a constant. Here, $a_1$ is called the first term, and the constant $d$ is called the common difference. An arithmetic sequence is also called an Arithmetic Progression (A.P.).
Common Difference of an Arithmetic Progression
An AP maintains a constant difference between consecutive terms, known as the common difference. For terms $a_1,a_2,a_3,a_4,a_5,$ and $a_6$ in an AP, the common difference can be expressed as $D=a_2−a_1=a_3−a_2=a_4−a_3=…$
nth Term of Arithmetic Progression (AP)
The nth term of an AP is given by the formula, $a_n=a+(n−1)d$
Where
a = First term of the sequence.
n = Term's position in the sequence.
d = Common difference.
Sum of First n Terms in an Arithmetic Progression (A.P)
The sum of the first 'n’ terms in an AP is calculated using the formula, $S_n=\frac n2[2a+(n−1)d]$
Where
Sn denotes the sum of the terms.
'n' is the number of terms being summed.
'a' is the first term.
'd' stands for the common difference.
NCERT Solutions for Class 10 Maths: Chapter Wise
Access all NCERT Class 10 Maths solutions from one place using the links below.
Also, read,
- NCERT Exemplar Class 10 Maths Solutions Chapter 5 Arithmetic Progressions
- NCERT Notes Class 10 Maths Chapter 5 Arithmetic Progressions
NCERT Exemplar solutions: Subject-wise
After completing the NCERT textbooks, students should practice exemplar exercises for a better understanding of the chapters and clarity. The following links will help students find exemplar exercises.
NCERT Books and NCERT Syllabus
Students should always check the latest NCERT syllabus before planning their study routine. Also, some reference books should be read after completing the textbook exercises. The following links will be very helpful for students for these purposes.
Frequently Asked Questions (FAQs)
Q: What is an Arithmetic Progression (AP)?
A:
An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This difference is called the common difference (d).
Q: What is the general form of an AP?
A:
The general form is:
a, a+d, a+2d, a+3d,…............................
where a = first term, d = common difference.
Q: What types of problems are asked from Class 10 Maths Chapter 5 Arithmetic Progressions?
A:
Here are the types of problems asked in this chapter.
Finding the nth term of an AP
Finding the sum of n terms
Word problems on daily life situations (e.g., seats in a stadium, savings, distances, etc.)
Q: Why is this chapter important for board exams?
A:
This chapter usually carries 6–8 marks in Class 10 Board Exams, often in the form of direct problems as well as application-based word problems.
Q: Are Arithmetic Progressions useful in real life?
A:
Yes, APs are used in daily life situations such as:
Calculating savings with fixed deposits every month
Arranging seats in a stadium/theatre
Distribution of items in patterns
Predicting growth in steps
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On Question asked by student community
Have a question related to CBSE Class 10th ?
Hello, according to CBSE and most Indian education boards, the typical age for appearing in the Class 10 exam is around 15-16 years. However, there are some exceptional cases where academically advanced students complete their studies early. If you have officially cleared Class 10 from CBSE, received a valid marksheet, and your school and board have accepted your age proof during registration, your certificate is considered valid. Problems only arise if your admission to Class 10 violated the board’s eligibility rules (for example, skipping multiple classes without approval). In such rare cases, the board might question the validity. But if CBSE has issued you the certificate after proper procedures, it is legally recognized for further studies, competitive exams, and jobs. If you are worried about age restrictions for certain exams (like 17 years minimum for NEET), you might need to wait until you meet the required age.
Thank you, and best of luck for your studies.
Hello,
Yes, you are right. To get eligible for the 10th board examination, you need a minimum of 75% attendance. This is a rule that applies to the CBSE, or Central Board of Secondary Education, recognized schools. There are exceptions to this rule in the scenario, like
- Medical Emergencies
- Personal Tragedies
- Official Sports Participation
I hope it will clear your query!!
hello,
These are the documents to take admission in 11th standard at your college from cbse to Maharashtra State Board.
- Class X Marksheet
- School Leaving Certificate
- Date of Birth Certificate
- Passport-Sized Photograph
- Migrant Certificate
- Aadhar Card or Government ID
I hope it resolves your query!!
To transfer from a CBSE school to a Maharashtra state board junior college for 11th standard, you'll need to provide several documents.
These typically include: your Class 10th mark sheet (original and attested copies), School Leaving Certificate/Transfer Certificate (TC) from your previous school, and a migration certificate if you're changing from a different board.
You'll also need a birth certificate, passport-sized photographs, and potentially an eligibility certificate from the Maharashtra State Board
The specific documents required may vary depending on the college and region. Verify the exact requirements with the college's admission office or official website.
Hello Aditya,
The documents required to take admission in 11th standard from CBSE to Maharashtra State Board are:
-
10th Marksheet (CBSE Board)
-
School Leaving Certificate (from previous school)
-
Migration Certificate (CBSE Board)
-
Aadhaar Card (student’s ID proof)
-
Passport size photographs
-
Caste Certificate (if applicable)
-
Domicile Certificate (if required by the college)
Some colleges may ask for additional documents. It’s best to check with the specific college for exact requirements.
Hope it helps !
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