Problems based on series are asked a number of times in competitive exams like SSC CGL,CHSLE,Banking and various management exams.So,it becomes necessary for us to learn this topic.In this post we learn short Tricks along with Traditional method of solving such problems.
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| Arithmetic Progression:- Quantities are in Arithmetic Progression (AP) when they increase or decrease by a common difference. |
| Now consider the given series:-2, 6, 10, 14 ...............( Common Difference is 4 ) 10, 7, 4, 1 -2 ......... ( Common difference is -3 ) How to find common difference in AP ? We can find common difference in AP by substituting any term of the series from its next term. ex- In the above example the common difference is 6 - 2 = 4 n th term in Arithmetic Progression( A P ) :- Let 'n' be the number of terms and 'a' be the first term and d is the common difference then- nth Term = a + (n-1)d Sum of given number of terms in an Arithmetic Progression:- Let, a be the first term, d is common difference, n be the total number of terms and l denotes the last term, then required Sum S can be found as:- S= n(a+1)/2 On, putting the value of last term in the equation it becomes:- S = n[2a + (n-1)d]/2 Some important points on Arithmetic Progression:- ** If any two terms of an AP is given the series can be determined by solving the two equations and finding the first term and the common difference. ** If three quantities of AP are given the middle quantity is called arithmetic mean. ** If three quantities a, b, c are in AP then their arithmetic mean will be given by :- b =(a+c)/2 Ex:- Find the 19 th term in the given AP 1, 4, 7, 10, . . . . . . . . . . . . Solution:- N th term of an ( arithmetic progression ) AP is given by nth Term = a + (n-1)d also common difference= 4 - 1 = 3 19 th term = 1 + ( 19 - 1) × 3 = 55 Alternate Quicker Method Add the common difference to the first term 18 times 19 th term = 1 + 3 × 18 = 55 Ex:- Finding Average of an AP 3, 7, 11, 15, 19, 23 Solution:- Traditional method for finding the average is- Average = (3+7+11+15+19+23)/6 = 13 Alternate Method:-If the given numbers are in AP then average can be given as: Average = (1st Term + Last Term)/2 = (3+23)/2 =13 Average = (2nd Term + 2nd Last Term)/2 = (7+19)/2 =13 Average = (3rd term + 3rd last term)/2 = (11+15)/2 = 13 Corresponding Terms in an AP :-In the above example Pair of terms ( 1 st and last terms ) , ( 2 nd term and second last term ), ( 3rd term and 3rd Last Term) these all terms are termed as Corresponding Terms in an AP Alternate method for finding the Sum of an AP:- Sum = Number of terms × Average Ex:- Find the Sum of 3, 7, 11, 15, 19, 23, 27, . . . . . . . 51 having 13 terms. Solution:-Since the given terms are in AP as 7 - 3 = 11 - 7 = 15 - 11 = 4 So, Sum = Number of terms × Average Average = (1st Term + Last Term)/2 = (3+51)/2 = 27 Sum = 13 × 27 = 351 Traditional Method for finding SUM of an AP:- S= n[2a +(n-1)d]/2 S= 13[2x3 +(13-1)x4]/2 = 351 How to find common difference of an AP, when two terms of an AP is given:- Ex:- If 3 rd term and 9 th term of an arithmetic progression is given, then find the common difference of the AP - , - , 11 , - , - , - , - , - , 35 Solution:- Moving from 3 rd term to 9 th term, we have to add common difference 6 times. Now, 35 - 11 = 24, so common difference will be Common Difference = 24/6 = 4 Ex:- Find the Sum of an AP which has 21 terms, where 3 rd term is 11 and 9 th term is 35. Solution:- Here from above example we can find common difference is 4, so Sum up to 21 terms of given AP = Number of terms × Average of the AP Here, 3 rd term is 11 and common difference is 4 So, 4 term is 11 + 4 = 15 5 th term = 15 + 4 = 19 So, Average = (11+19)/2 = 30/2 =15 Now Sum is given as = 21 × 15 = 315 |
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