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 .

Expert Answers
hala718 eNotes educator| Certified Educator

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

 

crmhaske eNotes educator| Certified Educator

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

krishna-agrawala | Student

Value of nth term of any arithmetic progression is given by the formula:

nth term = a(n) = a + (n-1)c

Where:

a = first term and

c =common difference

Using this equation:

a(1) = a

a(3) = a + 2c

a(4) = a + 3c

a(7) = a + 6c

It is given:

a(4) = 3*a(1)

Therefore:

a + 3c = 3a

== - 2a + 3c = 0   ...   (1)

Similarly as given:

a(7) = 2*a(3) + 1

Therefore:

a + 6c = 2*(a +2c) + 1

==> a + 6c = 2a + 4c + 1

==> -a + 2c = 1   ...   (2)

Multiplying equation (2) by -2

2a - 4c = -2   ... (3)

adding equation (1) and (3):

- 2a + 2a + 3c - 4c = -2

==> -c = -2

==> c = 2

Substituting this value of c in equation (2):

-a + 2*2 = 1

-a + 4 = 1

a = 3

giorgiana1976 | Student

We'll note the first term as a1 and the common difference as d.

Now, we'll put in relations the constraints from enunciation:

"The fourth term of an A.P. is equal to three times the first term":

a4  = 3a1

But a4 = a1 + 3d

a1 + 3d = 3a1

We'll combine like terms:

2a1 = 3d

We'll divide by 2 both sides:

a1 = 3d/2

 The second constraint from enunciation is:"the seventh term exceeds twice the third term by one"

a7 = 2a3 + 1

But a7 = a1 + 6d

a1 + 6d = 2a3 + 1, where a3 = a1 + 2d

a1 + 6d = 2a1 + 4d + 1

We'll combine like terms:

a1 = 2d-1

But a1 = 3d/2

3d/2  =2d - 1

3d = 4d - 2

d = 2

a1 = 3d/2 

a1 = 3*2/2

a1 = 3

neela | Student

T4=  3T1

T7 = 2T3+1 by data.

Solution:

T1 = a1.

T4 = a1+3d, Therefore T4 = 3a1, a1+3d = 3a1 , 3d = 2a1.....(1)

T7 = a1+6d = 2T3+1 = 2(a1+2d)+1

a1+6d = 2a1+4d+1

6d-4d  = 2a1-a1 +1

2d -1= a.........(2)

From (1) and (2),  3d = 2(2d-1) , d = 2.

From (2) a1 = 2(2)-1 = 3 .

d= 2

a = 3