Question

Find four consecutive terms of an A.P. whose sum is 12 and sum of 3rd and 4th term is

14. ( Assume four consecutive terms in an A.P. are a-d , a , a+d , a+ 2d)​

Answer 1

Answer:

The terms are -3, 1, 5, 9

Step-by-step explanation:

Given:

• Sum of 4 consecutive terms of the A.P is 12
• Sum of third term and fourth term is 14

To Find:

The four consecutive terms of the A.P

Concept:

Here we have to form linera equations in 2 variables and find the values accordingly.

Solution:

Let the first term be a - d

Let the second term be a

Let the third term be a + d

Let the fourth term be a + 2d

By given,

Sum of the 4 terms = 12

Hence,

a - d + a + a + d +a + 2d = 12

4a + 2d = 12

Divide the whole equation by 2

2a + d = 6

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

Now

The third term of the A.P = a + d

The fourth term of the A.P = a + 2d

By given,

Sum of 3rd and 4th term = 14

a + d + a + 2d = 14

2a + 3d = 14

Substitute the value of 2a from equation 1

(6 - d) + 3d = 14

6 + 2d = 14

2d = 14 - 6

2d = 8

  d = 8/2

 d = 4

Hence the common difference of the A.P is 4

Now substitute the value of d in equation 1

2a = 6 - 4

2a = 2

  a = 1

Now we have to find the terms of the A.P

First term = a - d

First term = 1 - 4

First term = -3

Second term = a = 1

Third term = a + d

Third term = 1 + 4

Third term = 5

Fourth term = a + 2d

Fourth term = 1 + 2 × 4

Fourth term = 1 + 8

Fourth term = 9

Hence the terms are : -3, 1, 5, 9

Verification:

Sum of terms = 12

-3 + 1 + 5 + 9 = 12

-2 + 14 = 12

12 = 12

Sum of 3r term and 4th term = 14

5 + 9 = 14

14 = 14

Hence verified.