12. Similarly find the 3rd, the10th and the nth terms. Therefore, and the given series doesn’t forms a A.P. Therefore , the series will be 10,20,30,40,50.. Therefore, this is not an A.P. \( -1,-\frac{1}{2}, 0, \frac{1}{2} \) 1. (iv) Given, -10, – 6, – 2, 2 … It will help you stay updated with relevant study material to help you top your class! (vi) 0.2, 0.22, 0.222, 0.2222 …. 8. because every term is 50 more than the preceding term. On subtracting equation (i) from (ii), we get. (xv) \(1^{2}, 5^{2}, 7^{2}, 73, \dots \), Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP, (I) \( a=7, d=3, n=8, a_{n}=? (A) 97 (B) 77 (C) −77 (D) −87, (ii) 11th term of the A.P. Given, the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. (v) a = – 1.25, d = – 0.25, \( a=10, d=10 \) And the common difference of these APs be d. Given that, difference between 100th term of the two APs = 100, a1−a2 = 100……………………………………………………………….. (i), Difference between 1000th terms of the two APs. These NCERT problems are solved by experts at BYJU’S. If you liked the video, please subscribe to our YouTube channel so that you can get more such interesting and useful study resources. \) 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. 16. (iv) -10, – 6, – 2, 2 … and their application in solving daily life problems. = 1-3=-2 series will be 10, 20, 30, 40, 50 …. (iii) a = 4, d = – 3 Clearly, it can be observed that the adjacent terms of this series do (ii) \( 2, \frac{5}{2}, 3, \frac{7}{2}, \ldots \) Solving these NCERT Solutions will help you understand the topic completely and help you lay a greater foundation for future studies. (v) \( 3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots \) Ans : \( a=10, d=10 \) Hence \( a_{n}=28 \) We know that Let −230 be the nth term of this A.P., and by the nth term formula we know. If 17th term of an A.P. Therefore, using nth term formula, we get. Find this value of x. Overview ... Class 10 Maths Notes are free and will always remain free. Therefore, we can write the given AP in reverse order as; and common difference, d = 248 − 253 = −5. Solution: 8. Since, an+1 – an or the common difference is not same every time. Therefore, and the given series doesn’t form a A.P. (x) Given that, l = 28,S = 144 and there are total of 9 terms. 5. Write first four terms of the AP, when the first term a and the common difference d are given as follows: Which of the following are APs? Therefore, we can see that these odd numbers are in the form of A.P. You can solve different types of questions with varying difficulty. (IV) \( a=-18.9, d=2.5, a_{n}=3.6, n=? 17. Let the series be \( a_{1}, a_{2}, a_{3}, a_{4} \dots \) Therefore, the A.P. Hence, n=10 Act as a basis to solve arithmetic progression problems asked in competitive examination. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. Therefore, there are 60 multiples of 4 between 10 and 250. 5.7). Therefore, d = 1/2 and the given series are in A.P. Let the series be \( a_{1}, a_{2}, a_{3}, a_{4} \dots \) Move Class 10 Maths Chapter 5 main page for … Therefore, this is an AP with common difference as 4 and first term as 7. (iii) Cost of digging for first metre = 150 Therefore, the missing terms are 53, 23, 8, and −7 respectively. \) Here, first term , a=0.6 5.5 Summary As per the given condition, eq. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. \( \begin{array}{l}{a_{n}=a+(n-1) d} \\ {0=-18+(10-1) d} \\ {18=9 d} \\ {d=\frac{18}{9}=2}\end{array} \) An arithmetic progression is a very basic and important topic to study as almost all the competitive exams will ask questions on arithmetic progression. Overview Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 Miscellaneous Exercise. Find the number of terms in each of the following A.P. series will be-1, -1/2, 0, 1/2. (iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre. \( \begin{array}{l}{=7+(8-1) 3} \\ {=7+(7) 3} \\ {=7+21=28}\end{array}\) will be 10, 20, 30, and 40. It also provides suitable examples which show different techniques to find the sum of the first n terms of AP. For more information on the study resources we provide, register with BYJU’S website or download BYJU’S Learning App for a customised learning experience. Therefore, 20th term from the last term of the AP 3, 8, 13, …, 253.is 158. Therefore, the contractor has to pay Rs 27750 as penalty. not have the same difference between them. In each stroke, the vacuum pump removes \(\frac{1}{4}\) of air remaining in the cylinder at a time. (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time. Here, first term \( a=\frac{1}{3} \) Let the series \( a_{1,} a_{2}, a_{3}, a_{4} \dots \) We know that, the formula for sum of nth term in AP series is, (ii) Given, −37, −33, −29 ,…, to 12 terms, (iii) Given, 0.6, 1.7, 2.8 ,…, to 100 terms, Common difference, d = a2 − a1 = 1.7 − 0.6 = 1.1, (iv) Given, 1/15, 1/12, 1/10, …… , to 11 terms, Common difference, d = a2 –a1 = (1/12)-(1/5) = 1/60, (ii) 34 + 32 + 30 + ……….. + 10

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