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| 1. |
In an ap the sum of first 10 terms is equal to the 100th term find the value of a |
| Answer» SUM of the first n terms is given BYS n = 2n [2a+(n−1)d]Putting n = 10, we getS 10 = 210 [2a+(10−1)d]210=5(2a+9d)2a+9d=42 ...............(1)Sum of the last 15 terms is 2565Sum of the first 50 terms - sum of the first 35 terms = 2565S 50 −S 35 =2565250 [2a+(50−1)d]− 235 [2a+(35−1)d]=256525(2a+49d)− 235 (2a+34d)=25655(2a+49d)− 27 (2a+34d)=51310a+245d−7a+119d=5133a+126d=513a+42d=171 ........(2)Multiply the equation (2) with 2, we get2a+84d=342 .........(3)Subtracting (1) from (3)d=4Now, SUBSTITUTING the value of d in equation (1)2a+9d=422a+9×4=422a=42−362a=6a=3So, the required AP is 3,7,11,15,19,23,27,31,35,39........ | |