Question

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 1

Answer:

The first term (a) is 3 & the common difference (d) is 2 .

Step-by-step explanation:

Given:

a4 = 3a …………...…(1)

a7 = 2a3 +1 …………..(2)

As we know that , an = a + (n -1) d

4th term :  

a4 = a + (4 - 1)d

a4 = a + 3d …………..(3)  

3rd term :  

a3 = a + (3 -1)d

a3 = a + 2d ………….(4)

7th term :

a7 = a +(7 - 1)d

a7 = a + 6d ……….(5)

On putting the value of eq 3 in eq 1

a4 = 3a

a + 3d = 3a

3a - a  = 3d

2a = 3d

a = 3d/2 ……………(6)

On putting the value of eq 4 & 5  in eq 2

a7 = 2a3 +1

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

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

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

2d = a + 1

2d = 3d/2 + 1

[From eq 6]

2d - 3d/2 = 1

(4d - 3d)/2 = 1

d/2 = 1

d = 1 × 2

d = 2  

On putting the value of d = 2 in equation 6  

a = 3d/2

a = (3×2)/2

a = 3  

Hence, the first term (a) is 3 & the common difference (d) is 2 .

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Answer 2

Answer:

First term, a = 3

Common difference, d = 2

Step-by-step explanation:

A.T.Q.

a₄ = 3a (1)

a₇ = 2a₃ + 1 (2)

As we know that

• an = a + (n - 1)d

Putting n = 3, we get

a₃ = a + (3 - 1)d

a₃ = a + 2d (3)

Putting n = 4, we get

a₄ = a + (4 - 1)d

a₄ = a + 3d (4)

Putting n = 7, we get

a₇ = a + (7 - 1)d

a₇ = a + 6d (5)

Now, putting eq(4) in eq(1), we get

→ a + 3d = 3a

Solving this equation, we get

3d = 3a - a

3d = 2a

a = 3d/2 (6)

Also, putting eq(3) and eq(5) in eq(2), we get

→ a + 6d = 2(a + 2d) + 1

Solving this equation, we get

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

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

2d = a + 1 (7)

Putting eq(6) in eq(7), we get

2d = 3d/2 + 1

2d - 3d/2 = 1

(4d - 3d)/2 = 1

d = 2

Putting this value in (6), we get

a = 3(2)/2

a = 3