Three-term Geometric Progression

Let the numbers be: $\frac{a}{b}, a, ab$ , which are in geometric progression. We have: $\frac{a}{b} \times a \times ab=a^3=17576$ . By taking cube roots, we get a=26.

If b is negative, then $\frac {a} {b}$ and $ab$ would be negative valves, hence the sum of the three terms would be less than or equal to $a$ , which is 26. This would contradict the condition that $\frac {a}{b} + a + ab = 175$ . Thus, $b$ is positive, hence we can conclude that $\frac {a}{b}$ and $ab$ are the largest and smallest numbers, in some order. Thus, their sum is $175-26=149$ .

Let us suppose the three numbers are a, ar, ar^2
(where r denotes its usual meaning, common ratio) According to the question, a+ar+ar^2 = 175 and a . ar . ar^2 = 17576 Calculating the value of ar from the second equation , ar = 26 Now as we just need to find the value of the sum of the largest and smallest numbers , my claim is that whether the progression is increasing or decreasing , the middle term would have a certain value that lies between the values of the first and last term ( no matter whichever of them is largest and which one is smallest) . So substituting value of ar in the first eqn, a + ar^2 = 175 - ar = 175 - 26 = 149 which is the required answer.

If we take a/r, a and ar as the sequence, then product of these numbers is 175. Hence a^3 = 17576 and a = 26. Value of a/r + ar = a ( (1/r + r ) = 175 - a = 175 - 26 = 149. Therefore sum of smallest and largest numbers is 149.

Let the three numbers be a/b, a and ab , where a and b belong to real numbers and b is not equal to 0. Now, product

          a/b.a.ab = 17576
    =>    a^3 = 17576
    =>    a = cube root (17576)
    =>    a = 26

Now, sum

          a/b + a + ab = 175

We've to find the sum of first and the last number

          a/b + ab = 175 - a

Putting value of a = 26, we get

          a/b + ab = 175 - 26
    =>    a/b + ab = 149

Thus, the required answer 149.

The Wikipedia page for a geometric progression states that the general form of a geometric sequence is ${a}$ , ${ar}$ , ${ar^2}$ .... and that of a geometric series is ${a}$ , $+ {ar}$ , $+{ar^2}$ ......

Thus, ${a} * {ar} * {ar^2} = 17576.$

This simplifies into ${a^3r^3} = 17576$

or ${(ar)^3} = 17576$ .

Taking the cube root on both sides gives us ${ar} = \sqrt[3]{ 17576} = 26.$

By observation, ${ar} = 26$ is the second term of the geometric sequence ${a}$ , $+ {ar}$ , $+{ar^2}$ ......

The question asks for the sum of the largest and smallest numbers, i.e. ${a}+{ar^2}$ which is simply ${a}$ , $+ {ar}$ , $+{ar^2} - {ar}$ or

$175-26$ , which gives us $149$ .

Here a+b+c=175 and a b c=17576 As from property of GP we get b=26 so, a+b=149(sum of smallest number(a) and sum of largest number (c))

let the terms be a,b,c with common ratio r.

let a=k/r, for some real number k. then, b=ar =k/r*r = k

and, c=br =kr

so now, abc=17576 which means

(k/r)*(k)(kr)=17576

k^3=17576

k=cube root of 17576 k=26

since b=k

then b =26

so, a+b+c=175

a+c+26=175

a=c=149

Let the three real nos. in G.P. be, x/r, x, xr (r is the common ratio) x/r * x * xr = x cube = 17576 (given) hence, x= 26 now according to the given condition, 26/r + 26 + 26r = 175 therefore, 26/r +26r = 175 - 26 = 149 Hence the sum of the largest and the smallest number is 149

Let 3 terms be a, ar, ar^2. So, a* ar * ar^2 = 17576, => ar = 26. So a+ar+ar^2 = 175 (given). So a + ar^2 = 175 - 26 = 149

take three no's: a/r,a,ar. now product of these three is (a,3). as we are given value of product ,we can find value of a. as we have to find sum of largest & smallest no,therefore greater no is ar& smaller no is a/r. a/r+ar=175-a=175-26=149

Assume 1st Term=a ,2nd term=ar, 3rd Term=ar^2 Product of 3 Terms: a ar ar^2=17576 a^3r^3=17576 ar=26 Sum of 3 Terms: a+ar+ar^2=175 a+26+ar^2=175 a+ar^2=149 1st Term+3rd Term=149 Since they are all real numbers(positive), hence 1st term(smallest)&3rd term(biggest)

Let $x_1, x_2, x_3$ be these real numbers. We know that:

$x_n = x_1 \cdot q^{n-1}.$ Thus:

$x_2 = x_1 \cdot q$ , $x_3 = x_1 \cdot q^2$ . Now we have:

$x_1 + x_2 + x_3 = 175$ and $x_1 \cdot x_2 \cdot x_3 = 17576 = 26^3$ or: $x_1 + x_2 + x_3 = 175$ and $x_1 \cdot x_1 \cdot q \cdot x_1 \cdot q^2 = x_1^3 \cdot q^3 = 26^3$ or: $x_1 + x_3 = 175 - x_2$ and $x_1 \cdot q = 26 <=> x_2 = 26$ . From here: $x_1 + x_3 = 175 - 26 = 149$ .

Let the three numbers are a/r , a, ar. Therefore, a/r * a * ar = 17576 So, a= 26. Again, a/r + a+ ar =175 Since, a/r is the smallest number and ar is the largest number, therefore sum of a/r and ar is found (by putting the value of a), = 149

Let the three numbers in g.p. be a/r , a, ar where the common ratio is 'r'. now product=17576 so, a/r x a x ar =17576 or a^3=17576 or a = 26 sum = 175 so, a/r +a + ar = 175. now we want to find sum of first and third no. i.e. a/r + ar so, a/r +a + ar = 175 a/r +ar = 175-a = 175-26 = 149 which is the required answer

The geometric progression can be written in the form: $\frac{a}{r}, a, ar$ where $r$ is the common ratio. We know that $\frac{a}{r} \times a \times ar = 17576$ , so $a^3=17576$ and $a=26$ . We also know that $\frac{a}{r}$ and $ar$ are the largest and smallest, not necessarily in that order. We can just take the total sum, 175, and subtract 26, the middle number, to get the sum of the other two. Therefore, the sum of the largest and smallest is 149.

take the three numbers as 'a/r', 'a' and 'ar' the product of the numbers is a/r X a X ar = 17576, which is equal to a^3. Therefore a is equal to cube root of 17576 which is 26. sum of numbers is a/r + a + ar = 175 Therefore, the sum of a/r and ar = 175 -a = 175-26= 149

Let gp be a/r , a, a r then according to question, (a/r) a (a r)=17576 so, a=cube root of 17576=26 a/r +a+a r=175 so a/r + a r=175-a=149

a=26, a(1/r+1+r)=175 1/r+r=149/26 a(1/r+r)=149

Let the middle term of the sequence be a. Then the terms are a/r, a, ar. The product is then equal to a^3 and it is easy to check that 17576 = 26^3. Hence a = 26 and since we are only concerned with the sum of the smallest and the largest of the three, we arrive at 175 - 26 = 149.

$a + ar + ar^2=175$

$(a)(ar)(ar^2)=a^3r^3=17576$

This means $ar=\sqrt[3]{17576}=26$

we want $a +ar^2$ and

we know $a + ar + ar^2=175$

so $a +ar^2=175-ar=175-26=\boxed{149}$

Let the first term of the expression be A and the common ratio be R.

Hence, we have A + AR + AR^2 = 175

and A AR AR^2 = (AR)^3 = 17576

or AR = 26.

Putting this value in the sum equation, the sum of the first and the last terms can be said to be

A + AR^2 = 175 - 26 = 149!

Let a , b and c be the numbers that forms the progression, being a the smallest and c the largest. Then we have:

• $a + b + c = 175$
• $a \times b \times c = 17576$

Let r be the common ratio of the sequence. Therefore,

• $b = a \times r$
• $c = a \times r^{2}$

And,

• $a + a \times r + a \times r^{2} = 175$
• $a \times a \times r \times a \times r^{2} = 17576$

Using the second equation, we have that:

• $a ^{3} \times r ^{3} = 17576$
• $\sqrt[3]{a^{3} \times r^{3}} = \sqrt[3]{17576}$
• $a \times r = 26$

Which is the value of the middle term, b . Hence:

• $a + 26 + c = 175$
• $a + c = \boxed{149}$