Answer :

Given,


4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1


a4 = 3a and a7 = 2a3 + 1


We know, an = a + (n – 1)d where a is first term or a1 and d is common difference and n is any natural number


When n = 4:


a4 = a + (4 – 1)d


a4 = a + 3d


When n = 7:


a7 = a + (7 – 1)d


a7 = a + 6d


When n = 3:


a3 = a + (3 – 1)d


a3 = a + 2d


According to question:


a7 = 2a3 + 1


a + 6d = 2(a + 2d) + 1


a + 6d = 2a + 4d + 1


a – 2a + 6d – 4d = 1


-a + 2d = 1


a – 2d = -1……(i)


a4 = 3a


a + 3d = 3a


3d = 3a – a


3d = 2a



Put this value of d in equation (i):






a = 3


Now, put this value of a in equation (ii):




d = 2


Hence, the value of a and d are 3 and 2 respectively


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