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Determine the AP whose third term is 16 and 17th term exceeds the 5th term by 12 |
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Answer» Here A3= a+2d=16.....(1)A17-A5=(a+16d)-(a+4d)=12or 12d=12Hence d=1from (1)a+2=16, a=14Thus the series is 14,15,16,17,18.......\xa0 Hope so you can understand... if you are having any doubt regarding this question so please ask to me thankuuu A3=16 a+2d=16a=16 - 2dWe have got the value of a Again A.T.Q A17 - A5= 12 a+16d - (a+4d) =12 On putting the value of a=16 -2d 16 - 2d + 16d - 16 - 2d - 4d = 12 16 will be cancel...So, -2d +16d -2d - 4d= 12 -4d +12d = 12 8d=12 d= 12/8 d=3/2 We have got the value of d=3/2 Now putting the value of d into the equation of a a =16 -2 ×3/2 a= 16 -3 a = 13..So we have got the value of both a and d... 4,10,16,22,28 |
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