class 10 maths chapter 5

By going through the summary part you can cover the entire chapter in a few points which help in memorizing the essential concepts. (ix) 1, 3, 9, 27 … If 17th term of an A.P. We can see that the cost of these prizes are in the form of A.P., having common difference as −20 and first term as P. Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40. Four terms of this A.P. Let the series \(a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \ldots \) When we divide 250 by 4, the remainder will be 2. (xiv) 12, 32, 52, 72 … (i) 2, 7, 12 ,…., to 10 terms. Taxi fare for first 4 km = 31 + 8 = 39 Cost of digging for first 4 metres = 250 + 50 = 300 We can see, that the given penalties are in the form of A.P. Write first four terms of the A.P. Determine the A.P. Here, first term, a = —5 and their application in solving daily life problems. (iv) \(a=-1, d=\frac{1}{2} \) Given, the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Ans : (i) It can be observe that (xiv) \( \mathrm{1}^{2}, 3^{2}, 5^{2}, 7^{2}, \ldots \) Therefore, the contractor has to pay Rs 27750 as penalty. The sum of the third and the seventh terms of an AP is 6 and their product is 8. (i) 2, 4, 8, 16 … Find the first three terms of the A.P. NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions – Here are all the NCERT solutions for Class 10 Maths Chapter 5. 2. If in the nth week, her weekly savings become Rs 20.75, find n. Given that, Ramkali saved Rs.5 in first week and then started saving each week by Rs.1.75. Solution: On subtracting equation (i) from (ii), we will get here. (ii) a = –2, d = 0 Therefore, putting the given values, we get. Clearly the series will be \( 1.25,-1.50,-1.75,-2.00 \ldots \ldots . Clearly, it can be observed that the adjacent terms of this series do Since, an+1 – an or the common difference is not same every time. It will help you stay updated with relevant study material to help you top your class! Hence, the sum of first two terms is 4. All these are divisible by 4 and thus, all these are terms of an A.P. \( =(-1)-(-5)=-1+5=4 \) Subtracting equation (i) from equation (ii), 10. \( \begin{array}{l}{=1.7-0.6} \\ {=1.1}\end{array} \), Ans : (I) \( a=7, d=3, n=8, a_{n}=? Thus, we can conclude now, that the rungs are decreasing in an order of AP. \) Therefore, we can see that these odd numbers are in the form of A.P. (ii) \(-5,-1,3,7 \ldots \) Therefore the series will be \( -2,-2,-2,-2 \ldots\) Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …, Common difference, d = Second term – First term, Common difference, d = a2 − a1 = 8 − 3 = 5. \( \begin{array}{l}{a_{1}=a=10} \\ {a_{2}=a_{1}+d=10+10=20} \\ {a_{3}=a_{2}+d=20+10=30} \\ {a_{4}=a_{3}+d=30+10=40} \\ {a_{5}=a_{4}+d=40+10=50}\end{array} \) (x) \( a, 2 a, 3 a, 4 a, \dots \) You can also download here the NCERT Solutions Class 10 Maths chapter 5 Arithmetic Progressions in PDF format. (iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre. find the common difference d and write three more terms. Thus we can see the houses numbered in a row are in the form of AP. Hence, the total volume of concrete required to build the terrace is 750 m3. 3. If the top and the bottom rungs are apart, what is the length of the wood required for the rungs? (i) and eq(ii) are equal to each other; As we know, the number of houses cannot be a negative number. The distances of potatoes from the bucket are 5, 8, 11, 14…, which is in the form of AP. Therefore, the A.P. \) 19. Chapter 5 - Set & Functions. For an A.P \( a_{n}=a+(n-1) d \) 3, 8, 13, 18, … is 78? Find the number of terms and the common difference. (iii) a = 4, d = – 3 An A.P. When we divide 999 by 7, the remainder will be 5. Solution: 8. What is the sum of first two terms? Keep visiting BYJU’S to get complete assistance for CBSE class 10 board exams. Therefore, volumes will be V, 3V/4 , (3V/4)2 , (3V/4)3…and so on. 200 for the first day, Rs. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. We know that , First four terms of this AP. Overview ... Class 10 Maths Notes are free and will always remain free. 3, 8, 13, …, 253. Now as we know, Your email address will not be published. (i) \( 2,4,8,16, \dots \) The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. After you have studied lesson, you must be looking for answers of its questions. 5.8). (iv) \( 0.6,1.7,2.8,3.9 \ldots \) In the following APs find the missing term in the boxes. As we can see from the figure, the ladder has rungs in decreasing order from top to bottom. Let us say the number of xth houses can be represented as; Sum of number of houses beyond x house = Sx-1, S49 – Sx = {49/2[2.1+(49-1)1]}–{x/2[2.1+(x-1)1]}. (v) a = – 1.25, d = – 0.25, Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …, Therefore, the A.P. (ii) − 37, − 33, − 29 ,…, to 12 terms because every term is 8 more than the preceding term Clearly, adjacent terms of this series do not have the same difference Four terms of this A.P. Let 10 be the nth term of this A.P., therefore, (iii) Given, (−5) + (−8) + (−11) + ………… + (−230). For an A.P \( a_{n}=a+(n-1) d \) Taxi fare for first 3 km = 23 + 8 = 31 \( \begin{array}{l}{a_{1}=a=4} \\ {a_{2}=a_{1}+d=4-3=1} \\ {a_{3}=a_{2}+d=1-3=-2} \\ {a_{4}=a_{3}+d=-2-3=-5}\end{array} \) . The first term of an AP is 5, the last term is 45 and the sum is 400.