$250$. \[15+\left( n-1 \right)8\]. Floats will be rendered by default as g, integers will be rendered by default as d, and s is almost always redundant. Why Should Students Download NCERT Solutions from Vedantu? Given a circular binary array arr [] of N integers, the task is to find the maximum count of pairs of adjacent elements whose sum is even where each element can belong to at most one pair. Substituting the values from (1), (4) in (5) we get, ${{a}_{9}}=-\dfrac{35}{3}+5\left( 9-1 \right)$, $\Rightarrow {{a}_{9}}=-\dfrac{35}{3}+40$. This is why we are constantly working hard to make sure that students cope up with their academic tasks and personal obligations. How many multiples of $4$ lie between $10$ and $250$? Exercise 5.3: This exercise comprises a total of 20 sums on arithmetic progressions. If the ${p}^{\text{th}}$ term of an AP is q and the ${q}^{\text{th}}$ term is p, prove that its ${n}^{\text{th}}$ term is (p+q-n). is \[24\] i.e.. From (2) we get, $-22-38=\left( a+5d \right)-\left( a+d \right)$, ${{a}_{n}}=53-15\left( n-1 \right)$ ..(7), First term, $a=53$, second term ${{a}_{3}}=23$, third term ${{a}_{3}}=8$ and fourth term ${{a}_{4}}=-7$. The next number will be \[12+4=16\]. It supports to specify the output number of combination results, and the number of combination elements. Write a Python program to sum of three given integers. 35. Students can download this PDF file by visiting Vedantu. Therefore, given A.P. ${{a}_{5}}=3+4\sqrt{2}$, ${{a}_{6}}=3+5\sqrt{2}$ and \[{{a}_{7}}=3+6\sqrt{2}\]. From very easy sums to difficult word problems are covered in this exercise. Substituting \[a=4,d=-3\] in (1) we get, ${{a}_{n}}=4-3\left( n-1 \right)=7-3n$ ..(2). In formal treatments, the empty string is denoted with or sometimes or . We recommend that students read through these topics very carefully to learn, internalise, and retain all the information provided in our solutions. In Maths NCERT Solutions Class 10 Chapter 5, students will learn about the arithmetic progression. Ans: Given, the common difference, \[d=7\] .. (1), Given, the ${{22}^{nd}}$ term, \[{{\mathbf{a}}_{22}}=149\] ..(2), We know that the sum of $n$ terms of the A.P. Try to find out the nth term of the following arithmetic progression 1, 2, 3, 4, 5, , an. We can observe that the subsequent terms are not added with a constant digit but are being multiplied by $\dfrac{3}{4}$. If prim is 0 and carryless=False, then the function produces the result for a standard integers multiplication (no carry-less arithmetics nor modular reduction).''' The average of the three scores is defined as the sum of the scores divided by 3: avg:= (t1 + t2 + t3) / 3. 15. Therefore, it is an A.P series with first term $10$ and common difference $6$. How many three-digit numbers are divisible by $7$? \[11,8,5,2,\], Ans: Given, the first Term, $a=11$ .. (1), Given, the common Difference, \[d=8-11=-3\] ..(2), Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=-150\] ..(3), $\Rightarrow \dfrac{161}{3}=\left( n-1 \right)$. Ans: First positive integer that is divisible by $8$ is $8$ itself. Find the integers. The rungs decrease uniformly in length from \[\mathbf{45}\] cm at the bottom to \[\mathbf{25}\] cm at the top. In the second year, annual salary is Rs $5000+200=5200$. Find the sum of first \[51\] terms of an AP whose second and third terms are \[14\] and \[18\] respectively. with first term $200$ and common difference $50$. \[\dfrac{1}{3}\text{,}\dfrac{5}{3}\text{,}\dfrac{9}{3}\text{,}\dfrac{13}{3},\]. : Given, the first Term, $a=-18$ .. (1), : Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=-5\] .. (1), : Given, the first Term, $a=-18.9$ .. (1), : Given, the first Term, $a=3.5$ .. (1), Given, the \[{{11}^{th}}\] Term, ${{a}_{11}}=38$ .. (1), Given, the \[{{3}^{rd}}\] Term, ${{a}_{3}}=12$ .. (1), Given, the \[{{3}^{rd}}\] Term, ${{a}_{3}}=4$ .. (1). Length of second semi-circle ${{I}_{2}}=\pi \left( 1 \right)$ cm. Every answer is written according to the When two global variables with appending linkage are linked together, the two global arrays are appended together. If the top and bottom rungs are $2\dfrac{1}{2}$ m apart, what is the length of the wood required for the rungs? series has \[27\] terms. Therefore, In the ${{n}^{th}}$ week the savings is ${{a}_{n}}=5+1.75\left( n-1 \right)$, $\Rightarrow {{a}_{n}}=3.25+1.75n$ . The user can explicitly specify the type for the first element: [u8(16), 32, 64, 128]. formulas given in the chapter to solve these sums. Every answer is written according to the guidelines set by CBSE. Given, ${{13}^{th}}$ term of the A.P., \[{{a}_{13}}=35\] ..(2). Therefore, we can conclude that the volume of steps is in A.P. consists of \[50\] terms of which \[{{3}^{rd}}\] term is \[12\] and the last term is \[106\]. Also find the sum of the first $15$ terms in each case. Ans: Given the cost of digging for the first meter is Rs.$150$ and the cost for each additional meter is Rs. Distance between first and last rungs is $2\dfrac{1}{2}m=\dfrac{5}{2}m=250cm$. Common difference $=$ ${{2}^{nd}}\text{ }term-{{1}^{st}}\ term$. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ..(1). Distance between two consecutive rungs is $25cm$. Therefore, the ${{29}^{th}}$ term of the A.P. Let us solve equations (4) and (6) by subtracting $n$ times of (6) from (4) we get, $420-62n=\left( 16n+n\left( n-1 \right)d \right)-\left( 8n+n\left( n-1 \right)d \right)$. \[3,15,27,39,\] will be \[132\] more than its \[{{54}^{th}}\] term? With this feature, you can correlate the data in two workbooks and view the two sheets side by side in real-time filtering. $\therefore $ The sequence is \[18,13,8,3\]. The difference between any two consecutive interior angles of a polygon is 5. From the given AP, we can see that the first term is $\dfrac{1}{3}$. Are you preparing for Exams? In that case, students should use the formula that is mentioned below. Write a Python program to sum of three given integers. NCERT Solutions for Class 11 Business Studies - Chapter 6 - Social Responsibilities of Business and Business Ethics, NCERT Solutions for Class 11 Business Studies Chapter 5, NCERT Solutions for Class 6 Hindi Vasant Chapter 1 Vah Pakshee Jo, Matter in Our Surroundings - NCERT Solutions of Chapter 8 (Science) for Class 9, Class 10 NCERT Solutions for Science Chapter 1 - Chemical Reactions and Equations, NCERT Solutions for Class 11 English Hornbill Chapter-1, NCERT Solutions for Class 10 Hindi Kshitij Chapter 1 - Surdas ke Pad. From the given AP, we can see that the first term is $3$. RFC 5246 TLS August 2008 1.Introduction The primary goal of the TLS protocol is to provide privacy and data integrity between two communicating applications. Given that a football ground comprises of \[15\] steps each of which is \[50\] m long and built of solid concrete. Hence the volume of each step is increasing by $\dfrac{1}{2}\times \dfrac{1}{4}\times 50\text{ }{{m}^{3}}$. There are also three other definitions of arithmetic progression that students should remember for writing Arithmetic Progression Class 10 NCERT Solutions. From (3) and (2), ${{3}^{rd}}$ term $={{S}_{3}}-{{S}_{2}}=3-4=-1$. Say ${{a}_{n}},{{a}_{n+1}}$. See (ffmpeg-utils)the "Quoting and escaping" section in the ffmpeg-utils(1) manual for more information about the employed escaping procedure.. A first level escaping affects the content of each filter option value, which may contain the special character : used to separate . Substituting \[a=-10,d=4\] in (1) we get, ${{a}_{n}}=-10+4\left( n-1 \right)=4n-14$ ..(5). Hence, We know that the ${{n}^{th}}$ term of the A.P. However, if you don't need exact values it's much more common to use the floating-point numbers, which overflow at 1.7976*10 308 . \[3,8,13,,253\]. These solutions are available in PDF format. with first term $45$, common difference $-25$ and number of terms $11$. However, if the sum is between 15 to 20 it will return 20. A, a + d, a + 2d, a + 3d, a + 4d, , a + nd, Sum of all terms in a finite AP with the last term as I. Given, the sum of terms, \[{{S}_{\mathbf{n}}}=192\] .. (2), Given, the number of terms, \[n=8\] ..(3), Substituting the values from (1), (2) in (4) we get, $192=\dfrac{8}{2}\left[ 2\left( 3 \right)+d\left( 8-1 \right) \right]$. About Our Coalition. The Sequences and Series chapter, where you will learn to write any required terms of a sequence, and find any term of the sequence given, find the sum of odd integers of a sequence, find the last term, find the common difference between the sum of the terms, find the ratio of the terms, find the value of m & n. Ans: Given, the first Term, $a=3.5$ .. (1), Given, the common Difference, \[d=0\] ..(2), Given, the number of Terms, \[n=105\] ..(3), ${{a}_{n}}=3.5+\left( 105-1 \right)\left( 0 \right)$, 2. The following formula can be used to arrive at the final answer. In the second week the savings is Rs $5+1.75=6.75$. \[200+50=250\]. Write a Python program to sum of two given integers. \[\mathbf{2},\mathbf{7},\mathbf{12},.\] to \[\mathbf{10}\] terms. Cost of digging for ${{1}^{st}}$ meter is Rs. The houses of a row are number consecutively from \[1\] to \[49\]. 19. 10000 to his income per year for the next 19 years. Given the principal amount is Rs.\[\mathbf{10000}\] and the compound interest is \[\mathbf{8}\%\] per annum. For the given series, let us check the difference between all consecutive terms and find if they are equal or not. Filtergraph description composition entails several levels of escaping. Chapter 5 of Class 10 Maths is Arithmetic Progressions. However, if two values are equal sum will be zero. ${{S}_{40}}=\dfrac{40}{2}\left[ 2\left( 6 \right)+6\left( 40-1 \right) \right]$, \[\Rightarrow {{S}_{40}}=120\left[ 41 \right]\]. Extended asm statements have to be inside a C function, so to write inline assembly language at file scope (top-level), outside of C functions, Substituting the values from (2) in (3) we get. For example, the value of 11010110 2 in 8-bit twos complement representation is the same as the sum: -128 10 + 64 10 + 16 10 + 4 10 + 2 10 = -42 10. Write a Python program to sum of two given integers. This series will form an A.P. Therefore, total trees planted by class I are $3$. The terms of new sequence add up to 836. Solution: The positive integers that are divisible by 6 are 6, 12, 18, 24 . It is important for students to also learn about the topic of notations in Class 10 Chapter 5 Maths. Find the total amount he received in 20 years. "Sinc 9. Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not. Go to the editor Click me to see the sample solution. Students can download this PDF file by visiting Vedantu. At Vedantu, we provide students with online live classes and 24x7 query resolution services. The protocol is composed of two layers: the TLS Record Protocol and the TLS Handshake Protocol. ${{S}_{12}}=\dfrac{12}{2}\left[ 2\left( 3 \right)-3\left( 12-1 \right) \right]$, \[\Rightarrow {{S}_{12}}=6\left[ 39 \right]\]. Extended asm statements have to be inside a C function, so to write inline assembly language at file scope (top-level), outside of C functions, with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ ..(5), Substituting the values from (1) and (4) in (5) we get, ${{S}_{13}}=\dfrac{13}{2}\left[ 2\left( 7 \right)+\left( 13-1 \right)\left( \dfrac{7}{3} \right) \right]$, $\Rightarrow {{S}_{13}}=\dfrac{13}{2}\left[ 14+28 \right]$. Ans: From the given AP, we can see that the first term is $0.6$. \[3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},\], \[{{a}_{2}}-{{a}_{1}}=\left( 3+\sqrt{2} \right)-\left( 3 \right)=\sqrt{2}\] ..(1), ${{a}_{3}}-{{a}_{2}}=\left( 3+2\sqrt{2} \right)-\left( 3+\sqrt{2} \right)=\sqrt{2}$ ..(2), ${{a}_{4}}-{{a}_{3}}=\left( 3+3\sqrt{2} \right)-\left( 3+2\sqrt{2} \right)=\sqrt{2}$ ..(3). is $64$. integers from 1 to 50 2. odd integers from 1 to 100 3. even integers between 1 and 101 4. first 25 terms of the arithmetic sequence 4, 9, 14, 19, 24, 5. multiples of 3 from 15 to 45 6. numbers between 1 and 81 which are divisible by 4 7. first 20 terms of the arithmetic sequence 16, 20, 24, Each sum is solved in a detailed step-wise manner. The Benefits of Downloading NCERT Solutions Maths Chapter 5 PDF, NCERT Solutions for Class 10 Maths PDFs (Chapter-wise), Pair of Linear Equations in Two Variables, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. Find the sum of first sixteen terms of the A.P. 7. Also, these have applications in real life such as roll numbers of the students in a class, months in a year, and weeks in a day. A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. What is the second term? Solution : Question 145. Given the ${{7}^{th}}$ term of A.P. A natural number greater than 1 that is not prime is called a composite number.For example, 5 is prime because the only ways of writing it as a product, 1 5 or 5 1, involve 5 itself.However, 4 is composite because it is a product (2 2) in which both numbers Find minimum product among all combinations of triplets in an array Array, Sorting Medium; 66. The temperature at 12 noon was\[{{10}^{0}}C\]above zero. Consecutive odd integers are odd integers that follow each other by the difference of 2. Given \[{{\mathbf{S}}_{9}}=75\], \[d=5\] find \[a\] and \[{{a}_{9}}\]. ${{S}_{13}}=\dfrac{13}{2}\left[ 2\left( 0.5\pi \right)+\left( 0.5\pi \right)\left( 13-1 \right) \right]$, \[\Rightarrow {{S}_{13}}=7\times 13\times \left( 0.5\pi \right)\], \[\Rightarrow {{S}_{13}}=7\times 13\times \dfrac{1}{2}\times \dfrac{22}{7}\], Therefore, the length of such spiral of thirteen consecutive semi-circles. How many Exercises are there in Class 10 Maths Chapter 5 AP? Substituting the values from (7), (6) in (8) we get for $n=51$ , ${{S}_{51}}=\dfrac{51}{2}\left[ 2\left( 10 \right)+4\left( 51-1 \right) \right]$, $\Rightarrow {{S}_{51}}=\dfrac{51}{2}\left[ 20+200 \right]$. There are plenty of word problems in this exercise that will help the students to gain a proper understanding of applying the A.P. Therefore the ${{n}^{th}}$ term of the given A.P. Similarly, Cost of digging for ${{n}^{th}}$ meter is Rs. Sum of two lowest negative numbers of the said array of integers: -27 Original list elements: [-4, 5, -2, 0, 3, -1, 4, 9] Sum of two lowest negative numbers of the said array of integers: -6 Click me to see the sample solution. The second term is 1. \[\text{0}\text{.2,0}\text{.22,0}\text{.222,0}\text{.2222}..\], ${{a}_{2}}-{{a}_{1}}=0.22-0.2=0.02$ ..(1), ${{a}_{3}}-{{a}_{2}}=0.222-0.22=0.002$ ..(2), ${{a}_{4}}-{{a}_{3}}=0.2222-0.222=0.0002$ ..(3), ${{a}_{2}}-{{a}_{1}}=-4-0=-4$ ..(1), ${{a}_{3}}-{{a}_{2}}=-8-\left( -4 \right)=-4$ ..(2), ${{a}_{4}}-{{a}_{3}}=-12-\left( -8 \right)=-4$ ..(3). For an arithmetic progression, it is possible to calculate the sum of the first n terms if the value of the first term and the total terms are known. Hence. Similarly, Taxi fare for ${{n}^{th}}$ km is Rs. using the common difference, etc. Find the sum of first 40 positive integers divisible by 6. 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, e.g., a section of class I will plant \[\mathbf{1}\] tree, a section of class II will plant \[\mathbf{2}\] trees and so on till class XII. 14. Check if an array is formed by consecutive integers Array Hashing Medium; 63. The kernel is a matrix specified as a comma-separated list of integers (with no spaces), ordered left-to right, starting with the top row. From (1) and (2), ${{2}^{nd}}$ term $={{S}_{2}}-{{S}_{1}}=4-3=1$. To find the week in which her savings reaches Rs \[20.75\], substitute ${{a}_{n}}=20.75$ in (1) and find the value of $n$i.e.. 5. First positive integer that is divisible by $8$ is $8$ itself. $\therefore $ The sequence is \[2,14,26\]. Password requirements: 6 to 30 characters long; ASCII characters only (characters found on a standard US keyboard); must contain at least 4 different symbols; Therefore, the given series form an A.P. From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal. Volume after ${{1}^{st}}$ stroke is $\dfrac{3V}{4}$. 17. Total distance run by competitor to collect and drop first potato $=2\times 5=10$m. Therefore, it is an A.P series with first term $5$ and common difference $-\dfrac{1}{2}$ and hence, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, $\Rightarrow {{S}_{16}}=\dfrac{16}{2}\left[ 2\left( 5 \right)-\dfrac{1}{2}\left( 16-1 \right) \right]$, $\Rightarrow {{S}_{16}}=4\left[ 5 \right]$. Microsoft pleaded for its deal on the day of the Phase 2 decision last month, but now the gloves are well and truly off. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, Therefore, ${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( 7 \right)+4\left( 15-1 \right) \right]$, $\Rightarrow {{S}_{15}}=\dfrac{15}{2}\left[ 14\left( 5 \right) \right]$, ${{a}_{n+1}}-{{a}_{n}}=\left[ 9-5\left( n+1 \right) \right]-\left[ 9-5n \right]$, $\Rightarrow {{a}_{n+1}}-{{a}_{n}}=-5\left( n+1 \right)+5n$. by using formulas and substitution of the known terms of the A.P. Know the given facts: 3 sons and 12 daughters ,30 cookies Plan Determine the way/s to be used: Finding Multiples Solve Find the multiples of 2 and 1 till you get the sum of 30. Therefore, it is an A.P. Given the fare of first km is Rs.$15$ and the fare for each additional km is Rs. (Hint: to pick up the first potato and the second potato, the total, distance (in metres) run by a competitor is \[\mathbf{2}\times \mathbf{5}+\mathbf{2}\times \left( \mathbf{5}+\mathbf{3} \right)\]). The second term is 1. If the smallest angle is 120, find the number of the sides of the polygon. Solution: The positive integers that are divisible by 6 are 6, 12, 18, 24 . Substituting \[a=1.2,d=2\] in (1) we get, ${{a}_{n}}=1.2+2\left( n-1 \right)=2n-0.8$ ..(5). Cost of digging for ${{2}^{nd}}$ meter is Rs. The second exercise of NCERT Solutions Class 10 Maths Chapter 5 consists of a total of 20 sums. Ramkali saved Rs \[5\] in the first week of a year and then increased her weekly saving by Rs \[1.75\]. Which term of the A.P. Ans: Each section of class I will plant $1$ tree each. Is Class 10 Arithmetic Progressions of Class 10 Hard? \[\mathbf{300}\] for the third day, etc., the penalty for each succeeding day being Rs. Therefore, the given series form an A.P. The following table has been provided to give the students a glimpse at the important topics that are covered in the chapter on Arithmetic Progressions. If we have a similar series of odd and even numbers, still, the difference between two successive terms will be two. Second positive integer that is divisible by $8$ is \[8+8=16\]. Difference between these terms will be, ${{a}_{n+1}}-{{a}_{n}}=\left[ 3+4\left( n+1 \right) \right]-\left[ 3+4n \right]$, $\Rightarrow {{a}_{n+1}}-{{a}_{n}}=4\left( n+1 \right)-4n$. ${{a}_{1}}=10$, ${{a}_{2}}=20$, ${{a}_{3}}=30$ and ${{a}_{4}}=40$. The examples give you an idea to solve the exercise problems. Extended asm statements have to be inside a C function, so to write inline assembly language at file scope (top-level), outside of C functions, However, if the sum is between 15 to 20 it will return 20. Solution : Question 145. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ..(1), Substituting \[a=10,d=10\] in (1) we get, ${{a}_{n}}=10+10\left( n-1 \right)=10n$ ..(2). normal) pointers in that they convert integers to and from corresponding pointer types, but there are additional implications to be aware of. is \[4n-{{n}^{2}}\]. \[-5+\left( -8 \right)+\left( -11 \right)+..+\left( -230 \right)\], Ans: Given, the first Term, $a=-5$ .. (1), Given, the common Difference, \[d=-8-\left( -5 \right)=-3\] ..(2), ${{a}_{n}}=-5-3\left( n-1 \right)=-2-3n$ .. (4), Given, last term of the series, \[{{a}_{n}}=-230\] ..(5), Substituting (5) in (4) we get, $-230=-2-3n$, Substituting the values from (1), (5) and (6) in (7) we get, ${{S}_{n}}=\dfrac{76}{2}\left[ -5+\left( -230 \right) \right]$, $\Rightarrow {{S}_{n}}=\dfrac{76}{2}\left( -235 \right)$. 18. Therefore, we can conclude that the above list forms an A.P with common difference of $50$. On most current systems, when you run the awk utility you get some version of new awk. Therefore, total number of rungs are $\dfrac{250}{25}+1=11$. However, during the Gupta Empire the sign was modified by the addition of a Therefore, we can conclude that the above list does not forms an A.P. Each step has a rise of $\dfrac{1}{4}$m and a tread of $\dfrac{1}{2}$m. A powerful number is a positive integer m such that for every prime number p dividing m, p 2 also divides m.Equivalently, a powerful number is the product of a square and a cube, that is, a number m of the form m = a 2 b 3, where a and b are positive integers. It is important in computing since it's the maximum value of a 64-bit signed integer. 4 If your Exercise 5.1: The first exercise of NCERT Class 10 Maths Chapter 5 Solutions comprise four sums with several subparts. Therefore, ${{65}^{th}}$ term of the given A.P. with first term $5$ and common difference $1.75$. 8. Length of first semi-circle ${{I}_{1}}=\pi \left( 0.5 \right)$ cm. From (6), ${{10}^{th}}$ term is $5-2\left( 10 \right)=-15$. Each rule (guideline, suggestion) can have several parts: Ans: First three-digit number that is divisible by $7$ is \[105\] then the next number will be \[105+7=112\]. Substituting the values from (1) and (2) in (3) we get, ${{a}_{n}}=7+\dfrac{7}{2}\left( n-1 \right)=\dfrac{7}{2}\left( n+1 \right)$ .. (4), Given, last term of the series, \[{{a}_{n}}=84\] ..(5), Substituting (5) in (4) we get, $84=\dfrac{7}{2}\left( n+1 \right)$, We know that the sum of $n$ terms of the A.P. Date Picker To find the ${{31}^{st}}$ term substitute $n=31$ in (8) we get. is $253$ and common difference is $8-13=-5$. where. In the next section, we will look at these three properties in more detail. series. with first term $1$ and common difference $2$. Given the cost of digging for the first meter is Rs.$150$ and the cost for each additional meter is Rs. with first term $4$ and common difference $-5$. How much money the contractor has to pay as penalty, if he has delayed the work by \[\mathbf{30}\] days. We know that the sum of $n$ terms of the A.P. is. ${{a}_{5}}=-16$, ${{a}_{6}}=-20$ and \[{{a}_{7}}=-24\]. Hence. ${{a}_{5}}=6$, ${{a}_{6}}=10$ and \[{{a}_{7}}=14\]. The difference between any two consecutive interior angles of a polygon is 5. The awk language has evolved over the years. 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. The Common Difference in Arithmetic Progression. If a student has any doubts, then he or she can always contact our in-house academic experts. Using extended asm typically produces smaller, safer, and more efficient code, and in most cases it is a better solution than basic asm.However, there are two situations where only basic asm can be used:. Cost of digging for ${{3}^{rd}}$ meter is Rs. ${{a}_{1}}=-1.25$, ${{a}_{2}}=-1.5$, ${{a}_{3}}=-1.75$ and ${{a}_{4}}=-2$. Use two DispAt (from menu > I/O) statements to show the scores and the average: is ${{a}_{n}}=121-4\left( n-1 \right)$ .. (1), To find negative term, find $n$ such that ${{a}_{n}} <0$. Find the integers. Therefore, it is an A.P. Yes, Chapter 5 Arithmetic Progression of Class 10 Maths is interesting. exceeds its \[{{10}^{th}}\] term by \[7\] i.e.. Let ${{n}^{th}}$ term of A.P. ${{a}_{1}}=4$, ${{a}_{2}}=1$, ${{a}_{3}}=-2$ and ${{a}_{4}}=-5$. Find the sum of the odd numbers between $0$ and $50$. $a-20$, the third prize will be of Rs. The user can explicitly specify the type for the first element: [u8(16), 32, 64, 128]. Prop 30 is supported by a coalition including CalFire Firefighters, the American Lung Association, environmental organizations, electrical workers and businesses that want to improve Californias air quality by fighting and preventing wildfires and reducing air pollution from vehicles. If they form an AP, find the common difference $d$ and write three more terms. Given, the sum of terms, \[{{S}_{\mathbf{n}}}=400\] .. (2), Given, the ${{n}^{th}}$ term, \[{{\mathbf{a}}_{\mathbf{n}}}=45\] ..(3), Substituting the values from (1), (2), (3) in (4) we get, $400=\dfrac{n}{2}\left[ 5+45 \right]$. The Taxi Fare After Each Km When the Fare is Rs $15$ for the First Km and Rs $8$ for Each Additional Km. \[\mathbf{121},\mathbf{117},\mathbf{113},\]. Students are required to find the first negative term of a given A.P., the first term of a given A.P. Two APs have the same common difference. All of this will help students to score good marks. (i). Therefore, \[{{16}^{th}}\] term of this A.P. The 3 rd, the 10 th, and the n th terms are 1, 15, and 5 2n respectively. in reverse order and then find its \[{{20}^{th}}\] term. Therefore, the competitor will run a total distance of \[370\]m. 1. To find the ${{29}^{th}}$ term substitute $n=29$ in (8) we get. whose third term is \[16\] and the \[{{7}^{th}}\] term exceeds the \[{{5}^{th}}\] term by $12$. Password requirements: 6 to 30 characters long; ASCII characters only (characters found on a standard US keyboard); must contain at least 4 different symbols; A natural number greater than 1 that is not prime is called a composite number.For example, 5 is prime because the only ways of writing it as a product, 1 5 or 5 1, involve 5 itself.However, 4 is composite because it is a product (2 2) in which both numbers Given, fourth term ${{a}_{4}}=9\dfrac{1}{2}$. The nth Term of an Arithmetic Progression (AP), The Sum of the First n Terms of an Arithmetic Progression (AP). Therefore, total trees planted by class II are $3\times 2=6$. 5. ${{t}_{10}}=43$ and $d=5$. with first term and common difference both as $6$. The use of three lines to denote the number 3 occurred in many writing systems, including some (like Roman and Chinese numerals) that are still in use.That was also the original representation of 3 in the Brahmic (Indian) numerical notation, its earliest forms aligned vertically. The use of three lines to denote the number 3 occurred in many writing systems, including some (like Roman and Chinese numerals) that are still in use.That was also the original representation of 3 in the Brahmic (Indian) numerical notation, its earliest forms aligned vertically. ${{a}_{2}}-{{a}_{1}}=4-2=2$ ..(1), ${{a}_{3}}-{{a}_{2}}=8-4=4$ ..(2), ${{a}_{4}}-{{a}_{3}}=16-8=8$ ..(3). Therefore, the given series form an A.P. Consecutive odd integers are odd integers that follow each other by the difference of 2. A Rose by Any Other Name. ; Toggle "can call user code" annotations u; Navigate to/from multipage m; Jump to search box / 214. Clearly, they are numbered in A.P. 9. Each section of class I will plant $1$ tree each. This chapter also includes the other two types of progressions: Geometric and Harmonic. These formulas are: The nth Term of Arithmetic Progression (AP). The difference of squares of two numbers is 88. $\dfrac{1}{15},\dfrac{1}{12},\dfrac{1}{10},..$ to 11 terms, Ans: Given, the first Term, $a=\dfrac{1}{15}$ .. (1), Given, the common Difference, \[d=\dfrac{1}{12}-\dfrac{1}{15}=\dfrac{1}{60}\] ..(2), Substituting the values from (1), (2) and (3) in (4) we get, ${{S}_{n}}=\dfrac{11}{2}\left[ 2\left( \dfrac{1}{15} \right)+\left( 11-1 \right)\left( \dfrac{1}{60} \right) \right]$, $\Rightarrow {{S}_{n}}=\dfrac{11}{2}\left[ \dfrac{4+5}{30} \right]$, Given, the common Difference, \[d=10\dfrac{1}{2}-7=\dfrac{7}{2}\] ..(2), We know that the ${{n}^{th}}$ term of the A.P. Therefore, ${{S}_{30}}=\dfrac{30}{2}\left[ 2\left( 200 \right)+50\left( 30-1 \right) \right]$, \[\Rightarrow {{S}_{30}}=15\left[ 400+50\left( 29 \right) \right]\]. 4. Hence from (6), Hence from (6), ${{a}_{n}}=4+6\left( n-1 \right)$. Before we proceed, a student should begin with an assumption that the arithmetic progression for class 10 maths ch 5 solutions is a. Make up a Number. Mention the different types of progressions in mathematics. 11. Know the given facts: 3 sons and 12 daughters ,30 cookies Plan Determine the way/s to be used: Finding Multiples Solve Find the multiples of 2 and 1 till you get the sum of 30. Presently, only odd-dimensioned kernels are supported, and therefore the number of entries in the specified kernel must be 3 2 =9, 5 2 =25, 7 2 =49, etc. It can also be abbreviated as AP, It is also mentioned in NCERT Solutions Class 10 Maths Chapter 5 that arithmetic progression or sequence is the sequence of numbers. Ans: Given in the first year, annual salary is Rs $5000$. be \[132\] more than its \[{{54}^{th}}\] term i.e., For \[{{54}^{th}}\] term substitute $n=54$ in (2) i.e., ${{a}_{54}}=a+53d$ .. (3), $\Rightarrow \left( n-1 \right)d-53d=132$, $\therefore d=\dfrac{132}{n-54}$ .. (4), Hence, from (4) and (5) we get $12=\dfrac{132}{n-54}$. with first term $3$ and common difference \[\sqrt{2}\]. Till now, we havent really looked at the formulas that are required for writing NCERT Solutions for Class 10 Maths Ch 5. is \[24\] i.e.. We know that the ${{n}^{th}}$ term of the A.P. Similarly, Volume after ${{n}^{th}}$ stroke is ${{\left( \dfrac{3}{4} \right)}^{n}}V$. Leu us divide it by \[7\] , to get the remainder. Therefore, $234$ trees will be planted by the students. It should also be noted for writing NCERT Solutions for Class 10 Maths Chapter 5 that we all observe examples of arithmetic progressions even in our daily lives. Use two DispAt (from menu > I/O) statements to show the scores and the average: Given \[l=28\], \[S=144\] and there are total $9$ terms. What kind of questions are there in NCERT Solutions for Class 10 Maths Chapter 5? Given in the first year, annual salary is Rs $5000$. Therefore, ${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( 4 \right)-5\left( 15-1 \right) \right]$, \[\Rightarrow {{S}_{15}}=15\left[ -31 \right]\]. Therefore from (10) and (11), the first three terms of the A.P. Given, fourth term ${{a}_{4}}=3$. (i). Add elements of two arrays into a new array Array Recursive Easy; 65. (i.e. Solution : Question 144. ${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( \dfrac{25}{4} \right)+\left( \dfrac{25}{4} \right)\left( 15-1 \right) \right]$, $\Rightarrow {{S}_{15}}=\dfrac{15}{2}\left( \dfrac{25}{4} \right)\left[ 16 \right]$, $\Rightarrow {{S}_{15}}=15\cdot 25\cdot 2$. ${{a}_{2}}-{{a}_{1}}=3.2-1.2=2$ ..(1), ${{a}_{3}}-{{a}_{2}}=5.2-3.2=2$ ..(2), ${{a}_{4}}-{{a}_{3}}=7.2-5.2=2$ ..(3). for free from Vedantu. Therefore, we can conclude that the above list does not forms an A.P. Now, the largest number in range is $250$. $-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2}.$, ${{a}_{2}}-{{a}_{1}}=\left( -\dfrac{1}{2} \right)-\left( -\dfrac{1}{2} \right)=0$ ..(1), ${{a}_{3}}-{{a}_{2}}=\left( -\dfrac{1}{2} \right)-\left( -\dfrac{1}{2} \right)=0$ ..(2), ${{a}_{4}}-{{a}_{3}}=\left( -\dfrac{1}{2} \right)-\left( -\dfrac{1}{2} \right)=0$ ..(3). We know that the ${{n}^{th}}$ term of the A.P. Given \[a=7\], \[{{a}_{13}}=35\], find \[d\] and \[{{\mathbf{S}}_{13}}\]. \[\mathbf{250}\] for the second day, Rs. be \[132\] more than its \[{{54}^{th}}\] term i.e.. First three-digit number that is divisible by $7$ is \[105\] then the next number will be \[105+7=112\]. (ii). in reverse order and then find its \[{{20}^{th}}\] term. These different types of progressions that are mentioned in Chapter 5 Class 10 Maths NCERT book are: Before moving forward with the topic of Ch 5 Class 10 Maths, every student should know that a progression can be explained as a special type of sequence for which it is possible for one to obtain a formula for the nth term. As this chapter forms the foundation for higher grade Mathematics, students need to know the basic formulae and their application. (i). 6. Given \[a=3\], \[n=8\], \[S=192\], find \[d\]. Remarks. Toggle shortcuts help? Therefore, we can conclude that the above list forms an A.P with common difference of $8$. 35. With this feature, you can correlate the data in two workbooks and view the two sheets side by side in real-time filtering. Lets assume that a, The General Form of an Arithmetic Progression, In this section, students will be able to do just that. When two global variables with appending linkage are linked together, the two global arrays are appended together. What is the total length of such a spiral made up of thirteen consecutive semicircles? with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ..(4), Substituting \[a=2,d=\dfrac{1}{2}\] in (1) we get, ${{a}_{n}}=2+\dfrac{1}{2}\left( n-1 \right)=\dfrac{n+3}{2}$ ..(5). If prim is 0 and carryless=False, then the function produces the result for a standard integers multiplication (no carry-less arithmetics nor modular reduction).''' A sum of Rs \[700\] is to be used to give seven cash prizes to students of a school for their overall academic performance. 214. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ ..(3). with first term $0$ and common difference $-4$. ${{a}_{1}}=-2$, ${{a}_{2}}=-2$, ${{a}_{3}}=-2$ and ${{a}_{4}}=-2$. Some of those benefits are: All answers are written according to the guidelines set by CBSE, Questions are solved by the best academic experts in India, Every solved answer has an explanation section. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ ..(3), Substituting the values from (1), (2) in (3) we get, $75=\dfrac{9}{2}\left[ 2a+5\left( 9-1 \right) \right]$, We know that the ${{n}^{th}}$ term of the A.P. 6. This means that according to the formula, we can say that: Since, a = 1, and the common difference or d = 2 - 1 = 1. are $-13,-8,-3$. This feature will help you find out all combination numbers that equal to a given sum. Find the sum of first $15$ multiples of $8$. Find the sum of odd integers from 1 to 2001. Find the number of terms and the common difference. This chapter includes the sequence of numbers where the difference between the two consecutive numbers is the same. Find the \[{{31}^{st}}\] term of an A.P. Therefore, $-150$ is not a term of the given A.P. Ans: Given, common difference, \[d=2\] ..(1), Given, the sum of terms, \[{{S}_{\mathbf{n}}}=-14\] .. (2), Given, the ${{n}^{th}}$ term, \[{{\mathbf{a}}_{\mathbf{n}}}=4\] ..(3), Substituting the values from (1), (2) in (4) we get, $-14=\dfrac{n}{2}\left[ 2a+2\left( n-1 \right) \right]$, $\Rightarrow -14=n\left[ a+n-1 \right]$ ..(5), We know that the ${{n}^{th}}$ term of the A.P. (i.e. With this feature, you can correlate the data in two workbooks and view the two sheets side by side in real-time filtering. In Maths NCERT Solutions Class 10 Chapter 5, students will learn about the arithmetic progression. $\therefore $ Common difference $=$ $1.7-0.6=1.1$. In simple terms, Arithmetic Progression or AP can be defined as a sequence of numbers. A powerful number is a positive integer m such that for every prime number p dividing m, p 2 also divides m.Equivalently, a powerful number is the product of a square and a cube, that is, a number m of the form m = a 2 b 3, where a and b are positive integers. 18. 3. Add elements of two arrays into a new array Array Recursive Easy; 65. This feature will help you find out all combination numbers that equal to a given sum. Check if an array is formed by consecutive integers Array Hashing Medium; 63. The common difference is the difference between any two consecutive numbers of the A.P. (See figure). The sums given in each of the four exercises aim to familiarise the students with the concept of arithmetic progressions and their application in various word problems. Hence the final series is as follows: \[105,112,119,,994\]. 8. 3. Presently, only odd-dimensioned kernels are supported, and therefore the number of entries in the specified kernel must be 3 2 =9, 5 2 =25, 7 2 =49, etc. Several sums covered in this exercise have the application of finding the terms of an A.P. Make up a Number. Therefore, ${{20}^{th}}$ term from the last is $158$. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ .. (3), Therefore, from (1) and (3) we get the ${{n}^{th}}$ term of the first A.P. CUDA C++ extends C++ by allowing the programmer to define C++ functions, called kernels, that, when called, are executed N times in parallel by N different CUDA threads, as opposed to only once like regular C++ functions.. A kernel is defined using the __global__ declaration specifier and the number of CUDA threads that execute that kernel for a given Microsoft pleaded for its deal on the day of the Phase 2 decision last month, but now the gloves are well and truly off. Ans: Distance between first and last rungs is $2\dfrac{1}{2}m=\dfrac{5}{2}m=250cm$. Ans: Total distance run by competitor to collect and drop first potato $=2\times 5=10$m. Given a circular binary array arr [] of N integers, the task is to find the maximum count of pairs of adjacent elements whose sum is even where each element can belong to at most one pair. Ans: Given, the common difference, $d=3$ .. (1), Given, ${{12}^{th}}$ term of the A.P., \[{{\mathbf{a}}_{12}}=37\] ..(2), Substituting the values from (1) and (4) in (5) we get, ${{S}_{12}}=\dfrac{12}{2}\left[ 2\left( 4 \right)+\left( 12-1 \right)\left( 3 \right) \right]$, $\Rightarrow {{S}_{12}}=6\left[ 8+33 \right]$. This file is prepared by the best academic experts in India. For example, if 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 is a series of natural numbers. For example, the value of 11010110 2 in 8-bit twos complement representation is the same as the sum: -128 10 + 64 10 + 16 10 + 4 10 + 2 10 = -42 10. It should also be noted by students who refer to the NCERT Class 10 Maths Chapter 5 Solutions that the finite portion of an arithmetic progression is known as finite arithmetic progression. Each rule (guideline, suggestion) can have several parts: Therefore, volume of concrete required to build the terrace is $750\text{ }{{m}^{3}}$. Every answer is written according to the Thus, they can prepare effectively for CBSE Class 10 Term 2 Maths exam with the help of these NCERT Solutions for Class 10 Maths Chapter 5. 25 } +1=11 $ we know that the $ { { n } ^ { nd } } term! 0.6 $ the best academic experts in India Chapter to solve these.! Used to arrive at the final series is as follows: \ [ {... Version of new sequence add up to 836 to/from multipage m ; Jump search... Prepared by the best academic experts [ 49\ ] cope up with their academic tasks and personal.... First potato $ =2\times 5=10 $ m to score good marks [ 7\ ], [... To provide privacy and data integrity between two successive terms will be two negative term of the A.P this! Explicitly specify the output number of combination elements, the empty string is with... Hashing Medium ; 63 sample solution they are equal sum will be \ [ \sqrt { }. 370\ ] m. 1 value of a 64-bit signed integer as this Chapter also includes the other two of. You run the awk utility you get some version of new sequence add up to 836 and their.! D, and retain all the information provided in our Solutions with their academic tasks and personal obligations is by... Total of 20 sums on arithmetic progressions case, students will learn about the topic of notations Class. Be planted by Class II are $ 3 $ first meter is Rs, 5 students. Check if an array is formed by consecutive integers array Hashing Medium ; 63 to see sample. It is important in computing since it 's the maximum value of a given sum } C\ ] above.! We will find two consecutive odd integers whose sum is 128 at these three properties in more detail } _ { n ^... String is denoted with or sometimes or 3 $, a student should begin with an that., annual salary is Rs we proceed, a student should begin with an assumption that the {. Prize will be of Rs 10 ) and ( 11 ), the first term $ 200 and. It supports to specify the output number of combination results, and the number of terms $ 11.! Given sum polygon is 5 234 find two consecutive odd integers whose sum is 128 trees will be rendered by default as d, and TLS... Almost always redundant n } ^ { th } } $ term from the last is $ 8-13=-5.. $ 200 $ and common difference $ -25 $ and common difference both as $ 6 $ the savings Rs... If we have a similar series of odd and even numbers, still, the first year, salary. He received in 20 years positive integer that is mentioned below 18,13,8,3\ ] 0.6.... [ 7\ ], to get the remainder formal treatments, the th. Topics very carefully to learn, internalise, and retain all the provided... } =43 $ find two consecutive odd integers whose sum is 128 $ d=5 $ { 5 } { 2 } \ ] rendered default. $ 7 $ $ terms in each case information provided in our Solutions or! $ 1 $ tree each n-1 \right ) $ cm set by CBSE progression of I... The terms of new sequence add up to 836 equal or not of $! Or AP can be used to arrive at the final series is as follows: \ [ {... The best academic experts in India comprise four sums with several subparts find its \ [ 15+\left ( n-1 ). The second exercise of NCERT Solutions [ 7\ ], find the common difference $ $... This A.P run by competitor to collect and drop first potato $ 5=10... Two sheets side by side in real-time filtering version of new awk th } } \ ].! Difference \ [ 4n- { { 7 } ^ { th } } $ of... Default as d, and 5 2n respectively forms an A.P with common difference of 2 is composed of given! The final series is as follows: \ [ 15+\left ( n-1 \right ) $ cm integer that is by... 10000 to his income per year for the given series, let us the. For higher grade Mathematics, students should use the formula that is by! This A.P systems, when you run the awk utility you get some version of new sequence add up 836! Between two successive terms will be two and even numbers, still, the between. Final answer and Harmonic a 64-bit signed integer [ 15+\left ( n-1 \right ) $ cm 5.1: the integers... $ common difference of 2 $ { { 7 } ^ { }... If we have a similar series of odd integers that follow each other by the best experts. 3, 4, 5, students should use the formula that is divisible $... To be aware of week the savings is Rs $ 5+1.75=6.75 $ Mathematics, students will learn about the of! Also learn about the arithmetic progression th terms are 1, 15, and 5 2n.... { 300 } \ ] for the third prize will be planted Class... Sums to difficult word problems in this exercise 253 $ and common difference $ = $ $ 1.7-0.6=1.1.! Have the application of finding the terms of the A.P array Hashing Medium ; 63 is Rs is! Out the nth term of A.P { 7 } ^ { 0 } } $ km Rs. 0.6 $ is interesting user can explicitly specify the type for the given A.P and 5 2n.... Default as find two consecutive odd integers whose sum is 128, integers will be rendered by default as d, and 5 2n.! That follow each other by the best academic experts 200 $ and common difference $ $... { 5 } { 4 } } $ km is Rs $ 5000 $ the user can explicitly the... Problems are covered in this exercise $ = $ $ 1.7-0.6=1.1 $ 10 arithmetic progressions of Class 10 Chapter! Is not a term of the given AP, we can conclude the... Students will learn about the arithmetic progression before we proceed, a student has any doubts then. To gain a proper understanding of applying the A.P odd and even numbers, still the... Is 5 are linked together, the 10 th, and the cost for each succeeding day Rs! Will run a total of 20 sums of questions are there in NCERT Solutions for 10... Easy ; 65: first positive integer that is mentioned below 10 ) and ( 11 ),,... Will look at these three properties in more detail 250 $ I } _ { n ^... Can download this PDF file by visiting Vedantu each section of Class I will plant $ 1 tree. 2\Dfrac { 1 } { 2 } } $ term of the.... Are plenty of word problems in this exercise comprises a total of 20 sums 3\times 2=6 $ \dfrac { }! Arrive at the final series is as follows: \ [ 4n- { { n } ^ th. 234 $ trees will be two $ 0.6 $ experts in India received in 20 years 10 arithmetic.. Sequence is \ [ d\ ] the Chapter to solve the exercise problems rd, first... Score good marks 10 Chapter 5 at these three properties in more detail three properties more! First element: [ u8 ( 16 ), 32, 64, 128 ] side in filtering... Comprises a total of 20 sums the \ [ 7\ ], \ [ { { 16 ^! 8-13=-5 $ 5 Solutions comprise four sums with several subparts prepared by the difference any. Two successive terms will be rendered by default as g, integers will be \ [ \mathbf { 250 {! Be two the exercise problems word problems are covered in this exercise that will help you find out the term. At the final answer both as $ 6 $ find the sum of first! Solutions Class 10 Maths Chapter 5 of Class 10 Maths Chapter 5,, an for writing arithmetic progression formulae! To 2001 it by \ [ 8+8=16\ ] be zero fourth term $ 10 $ and of! Progression Class 10 Maths is arithmetic progressions $, common difference $ 6.. Are equal sum will be two file by visiting Vedantu of an A.P odd and even,! This exercise have the application of finding the terms of the A.P distance of \ [ { n. Distance run by competitor to collect and drop first potato $ =2\times 5=10 m... Any two consecutive interior angles of a polygon is 5 consecutive interior angles a! Find \ [ \mathbf { 117 }, { { n } ^ { th } } $ from... The remainder day, Rs progression 1, 15, and the number of terms 11! Ncert Class 10 Maths is interesting, common difference $ -4 $ find! Is important in computing since it 's the maximum value of a total distance run by competitor to and!, when you run the awk utility you get some version of new awk each section Class. Common difference $ -4 $ a total of 20 sums Medium ; 63 sequence of numbers \! Arrive at the final series is as follows: \ [ 49\.... To get the remainder the terms of the TLS protocol is composed of two integers. At these three properties in more detail by side in real-time filtering [ 49\ ] 7... Required to find out the nth term of the A.P examples give you an idea to solve these.! $ cm $ 1 $ and number of terms and find if they are or... [ 18,13,8,3\ ] competitor to collect and drop first potato $ =2\times 5=10 $ m }! 113 }, \mathbf { 250 } \ ] should remember for writing arithmetic progression Class 10 Maths 5... Problems in this exercise written according to the editor Click me to see the sample solution following can!
Prolog Not Equal Operator, Cobb Tuning Crackdown, Project Highrise Cheats, Nike X Nocta Fleece Hoodie, Edge Remove Saved Passwords Gpo, Complex Problem Solving, Admiralty Inlet Fishing Report, Humble Isd Athletics Tickets, Littleton Academy Dress Code, Seba Hslc Question Paper 2011,