# need to calculate sum of 3 consecutive terms of arithmetic sequence x+4,3x,13x-2

The terms x + 4, 3x and 13x - 2 are consecutive terms of an arithmetic progression. So we know that they have a common difference

13x - 2 - 3x = 3x - x - 4

=> 10x - 2 = 2x - 4

=> 8x = -2

=> x = -1/4

So the three given terms are -1/4 + 4 = 15/4

3* -1/4 = -3/4

13*(-1/4) - 2 = -21/4

The sum of the three terms is 15/4 - 3/4 - 21/4 = -9/4

**The required result is -9/4**

If x+4,3x,13x-2, are the consecutive terms of an arithmetical progression, then the middle term is the arithmetical mean of x+4 and 13x-2.

3x=(x+4+13x-2)/2

3x =(14x-2)/2

By cross multiplying, the result will be:

6x=14x-2

We'll subtract 14x both sides:

6x-14x = -2

-8x = -2

We'll divide by -8:

x = 2/8

x = 1/4

The consecutive terms are:

x + 4 = 1/4 + 4 = (1+16)/4 = 17/4

3x = 3/4

13x - 2 = 13/4 - 2 = (13-8)/4 = 5/4

The sum of terms is:

x + 4 + 3x + 13x - 2 = 17x + 2

17x + 2 = 17/4 + 2

17x + 2 = (17+8)/4

**x + 4 + 3x + 13x - 2 = 17x + 2 = 25/4**

To find the sum of the 3 consecutive terms of an AP whose consecutive terms are x+4, 3x and 13x-2.

Since the 3 terms are consecutive, the common difference is:

3x-(x+4) = 13x-2-3x

=> 3x-x-4 = 10x -2

=> 2x-4 = 10x-2

=> 2-4 = 10x - 2x = 8x.

=> -2 = 8x.

So x = -2/8 = -1/4.

So the three consecutive terms are -1/4+4, -3/4 and 13(-1/4)-2

Or 15/4 , -3/4 and -21/4.

Sum of the terms = (15-3-21)/4 = -9/4

**So the sum of the three consecutive terms = -9/4.**