# The 4th</sup

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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