Q. 164.9( 7 Votes )

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

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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