The fourth term of an A.P. is equal to three times the first term and the seventh term exceeds twice the third term by one . Calculate the common difference and the first term .

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4th term os A.P = 3 times 1st

==> a4 = 3a1

a7 = 2*a3 + 1

let a1, a2, ...., an be an A.P

==> a1 = a1

a2 = a1 + r

a3 = a1 + 2r

a4 = a1+ 3r

a7 = a1 + 6r

We know that:

...

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4th term os A.P = 3 times 1st

==> a4 = 3a1

a7 = 2*a3 + 1

let a1, a2, ...., an be an A.P

==> a1 = a1

a2 = a1 + r

a3 = a1 + 2r

a4 = a1+ 3r

a7 = a1 + 6r

We know that:

a4 = 3a1

==> a1 + 3r = 3a1

==> 2a1 = 3r ....(1)

We also know thatL

a7 = 2*a3 + 1

a1 + 6r = 2(a1+2r) + 1

a1 + 6r = 2a1 + 4r + 1

==> 2r= a1 + 1 ......(2)

From (1) AND (2):

a1= (3/2)*r

Substitue in (2)

2r = (3/2)r + 1

==> (1/2)r = 1

==> r= 2

==> a1 = 3/2 * 2 = 3

a1= 3

a2 = 5

a3= 7

a4= 9

a5= 11

a6= 13

a7= 15

Let us verify:

a4 = 3a1

9= 3*3

9=9

Also:

a7= 2*a3 + 1

15 = 2*7 + 1

15 = 15

 

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According to the information provided:

a4 = 3a1
a7 = 2a3 +1

We want to dry and solve for d:

an = a1 + (n-1)d
a4 = a1 + (4-1)d
3a1 = a1 + (4-1)d
2a1 = 3d
a1 = (3/2)d

a7 = a1 + (7-1)d
a7 = (3/2)d + 6d = (15/2)d

a3 = a1 + (3-1)d
a3 = (3/2)d + 2d = (7/2)d

Using:
a7 = 2a3 +1

(15/2)d = 7d + 1
(1/2)d = 1
d = 2

Therefore a1 = 3

Approved by eNotes Editorial Team