How to find the common difference of AP when the first and last term are given Question: The first and last terms of an AP are 6.7 and 17.1 respectively. If there are 14 terms in the sequence, find its common difference. Solution First term of the sequence a = 6.7 Last term of the sequence Tn = 17.1 Since the sequence has 14 terms, it means that the 14th term is the last term Therefore: Tn = T14 = 17.1 Recall the nth term formula Tn = a + (n-1)d Substitute Tn for T14, a for 6.7 and n for 14 to get the formula below T14 = 6.7 + (14-1)d T14 = 6.7 + 13d Substitute T14 for 17.1 T14 = 6.7 + 13d 17.1 = 6.7 + 13d Subtract 6.7 from both sides to get 17.1 - 6.7 = 6.7-6.7 +13d 10.4 = 13d Divide through by 13 10.4/13 = 13d/13 d = 104/130 The common difference is 104/130 CHECK Substitute 104/130 for d in T14 = 6.7 + 13d T14 = 6.7 + 13*104/130 = (6.7*130 + 13*104)/130 = (871 + 1352)/130 ...
Tell us your mathematical challenge and we will be glad to assist you.