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