# What is x if 2, (3x+2), (24x+2) and 686 are consecutive terms of a GP.

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### 2 Answers

We are given that 2, (3x+2), (24x+2) and 686 are consecutive terms of a GP.

This means (3x + 2)/2 = (24x + 2)/ (3x + 2) = 686/ ( 24x + 2)

(3x + 2)/2 = 686/ ( 24x + 2)

=> (3x + 2)/2 = 343/ ( 12x + 1)

=> 686 = (3x +2)(12x +1)

=> 686 = 36x^2 + 24x + 3x + 2

=> 36x^2 + 27x – 684 = 0

=> 12x^2 + 9x – 228 = 0

=> 4x^2 + 3x – 76 =0

=> 4x^2 + 19x – 16x – 76 =0

=> x(4x + 19) – 4x(4x + 19) = 0

=> (4x + 19)(x – 4) =0

x = 4 and x = -19/4

But we find that only x = 4 is a valid root to satisfy the given GP.

**Therefore the required value of x is 4. **

Given that:

2, (3x+2), (24x+2) , 686 are terms of a geometric progression.

Let r be the common difference between terms.

Then we know that:

a1= 2

a2= 2* r = (2x+2)..............(1)

a3= 2*r^2 = 24x + 2...........(2)

a4= 2r^3 = 686.................(3).

We will solve for r in eq (3).

==> 2r^3 = 686

==> r^3 = 686/2 = 343

**==> r= 7**

Now we will solve for x in eq(1).

==> 2r= 3x + 2

2*7 = 3x + 2

==> 14 = 3x+2

==> 14 = 3x + 2

==> 3x = 12

**==> x = 4**

**Now we will write the terms.**

2, (2x+3) , (24x+2) , 686

==> 2, (3*4+2) , (24*4 + 2) , 686

==> 2, 14, 98, 686 are terms of a G.P where the common difference is r = 7