# The terms 3^x - 1 , 3^(2x + 2), 5*3^x + 1 are the terms of an a.p. Determine the value of x .

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3^x - 1 , 3^(2x+2) , 5*3^x + 1

Since the above are terms in an A.P, then:

3^(2x+2) - (3^x - 1) = 5*3^x + 1 - (3^(2x+2)

3^(2x+2) - 3^x + 1 = 5*3^x + 1 - 3^2(x+1)

Group Similar:

2*3^(2x+2)) - 6*3^x = 0

Divide by 2:

==> 3^(2x+2) - 3*3^x = 0

==> 3^(2x+2) = 3*3^x

==> 3^(2x+2) = 3^(x+1)

==> 2x + 2 = x+1

==> x= -1

We'll have to determine x, so that the given terms to be the consecutive terms of an arithmetic series.

Now, we'll apply the rule of the consecutive terms of an arithmetical progression. According to the rule, the middle term is the arithmetical mean of the joined terms.

3^(2x + 2) = ( 3^x - 1 + 5*3^x +1)/2

Eliminating like terms, from the brackets, from the right side, we'll get:

3^(2x + 2) = ( 3^x + 5*3^x)/2

Now, we'll factorize by 3^x, to the right side:

3^(2x + 2) = 3^x*(5+1)/2

3^(2x + 2) = 3*3^x

3^(2x + 2) = 3^(x+1)

Because the bases are matching, we'll apply the one to one property:

2x + 2 = x + 1

We'll isolate x to the left side:

2x - x = 1 - 2

x = -1

So, the consecutive terms of the A.P. are:

1/3 - 1 , 3^0, 5/3 + 1

**-2/3 , 1 , 8/3**

As 3^x-1, 3^(2x+2) and 5*3^x+1 are the consecutive terms of an arithmetic progression, we know that:

3^x-1+ 5*3^x+1= 2*3^(2x+2)

=>3^x+5*3^x=2*3^(2x+2)

=>3^x+5*3^x=2*3^2x*3^2

=>3^x+5*3^x=18*3^2x

=Let 3^x=y, now we have

y+5y=18y^2

=>6y=18y^2

=>y=3y^2

=>3y^2-y=0

=>y(3y-1)=0

=> y=0 or y=1/3

Now as 3^x=y

3^x=0 or 3^x=1/3

3^x=0 is not a valid solution. For 3^x=1/3, x=-1

**Therefore the valid value of x is -1.**

Since the terms are in AP, the common difference between the successive terms should be equal.

3^(2x+2) - (3^x -1) = 5*3^x+1 - 3^(2x+2)

9*3^2x -3^x +1 = 5*3^x +1 -9*3^2x

Collect 3^2x and 3^x together.

18*3^2x - (5+1)3^x -2

18t^2 -6t = 0, where t = 3^x and 3^2x = (3^x)^2 = t^2.

6t(3t-2) = 0

t = 0 or 3t-2 = 0

t = 0 gives: 3^x = 0 or x = -infinity.

3t -2 = 0 gives 3*3^x -2 = 0 or 3^x = 2/3. Or x = log(2/3)/ log3.