Hah! 3 things you have to solve for!

Given that the three successive terms in GP are $4,(17-x),(31+x),....$ with first term $(a)=4$ and the common ratio be $r$ . Now, we know that in a GP, $\large \frac{T_n}{T_{n-1}}=r$ . Using this fact, we have two equations for $r$ . They are ----

$\large r=\frac{17-x}{4}=\frac{31+x}{17-x}$ . So----

$\large \frac{17-x}{4}=\frac{31+x}{17-x}$

$\implies (17-x)^2=4(31+x)$

$\implies 289-34x^2+x^2=124+4x$

$\implies x^2-38x+165=0$

$\implies (x-5)(x-33)=0$

$\implies x-5=0\quad \text{ or }\quad x-33=0$

$\implies x=5 \quad \text{ or }\quad x=33$

Now, both the solutions of $x$ are positive, so we take both the solutions into account for now !

Now, from original equation of $r$ , we have ---

$r=\frac{17-x}{4} \implies 4r=17-x ......(i)$

Here, in eq. (i), if we put $x=33$ , this makes $r$ (-ve) which is in violation of the given criteria. So, $x\neq 33$ . Thus, we have $x=\boxed{5}$ . Putting this in eq. (i), we get-----

$4r=17-5 \implies 4r=12 \implies r=\boxed{3}$

Now, we have the values of $a,r$ and we should know that the $n^{th}$ term of a GP is calculated by the formula $a_n=a.r^{n-1}$ where $a_n$ represents the $n^{th}$ term of a GP.

So, the $5^{th}$ term in the series is equal to the value of $a.r^{5-1}=a.r^4$ . So----

$a_5=a.r^4=4\times 3^4 = 4\times 81 = \boxed{324}$

Now, the reqd. sum $=x+r+a_5 = 5+3+324=\boxed{332}$

Hello,1st,you must find the value of x,

By using common ratio,

(17 - x) / 4 = (31 + x ) / 17 - x

(17 - x)(17 - x) = 4(31+ x)

289 - 34x+ x^2 = 124 + 4x

x^2 - 38x + 165 = 0

(x-33)(x-5)=0

x=33,or x=5

when x=33,the sequence is 4,-16,64(r is -,no need to take the value) as r=-4=-16/4=64/-16

when x=5,the sequence is 4,12,36(take this one),r=12/4=36/12=3

5th term=a.r^4=4.(3)^4=324

Therefore,x+r+ 5th term= 5+3+324=332

Thanks...

r=(17-x)/4=(31+x)/(17-x) solve the equation... x=5 r=3 5th term of GP=4.r^(4) =324 sum=5+3+324=332

In order to find x, use r's definition in geometric progression:‎ ‎(17-x)/4= (31+x)/ (17-x)‎ Algebraic manipulation (particularly, cross multiplication) will result:‎ ‎0 = x^2 -38x + 165‎ which when factorized gives ‎0 = (x-33) (x-5)‎ and by the zero product property, x = 33 or x = 5.‎

Thusly,‎ r = (17-33)/4 or r = (17-5)/4‎ Equivalently,‎ r = -4 or r = 3‎

Here is where reading (or rereading) skills helps. Extracting from the question: "Where x and r are positive integers."‎

From the question, then,‎ r = 3.‎

Obtain "the fifth term of the sequence," from the general equation of geometric progression: T (n) = a (r) ^n-1‎, T (5) = 4 (3) ^5-1 = 4*(3) ^4 = 4(81) = 324‎

Referring back to the question, one tries to answer the question of the sum of x, r and T (5):‎ ‎5 + 3 + 324 = 332‎

First, we must find what the common ratio and $x$ are. For any $r$ , then the third term, which is $4 \times{r^{2}}$ , must be greater than 32, because $x$ is positive. So the only considerable option for $r$ is 3. 4 is obviously way too huge for $17-x$ to be 4 times 4, which is 16. If we plug 3 in for $r$ , and let $x$ be equal to 5, it works. So with that, we can find out that the fifth number in the sequence is 324. Adding $324+3+5$ , we get $\boxed{332}$ .