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8 - 9*r

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Q: 8 minus the product of 9 and r?
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What is 8 minus the product of 9 and r?

If we assume that "r" is unknown: -9r+8 or 8-9r It cannot be simplified further.


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I'll try to answer the question, "If the 5th term of a geometric progression is 2, then the product of its FIRST 9 terms is --?" Given the first term is A and the ratio is r, then the progression starts out... A, Ar, Ar^2, Ar^3, Ar^4, ... So the 5th term is Ar^4, which equals 2. The series continues... Ar^5, Ar^6, Ar^7, Ar^8, ... Ar^8 is the 9th term. The product P of all 9 terms is therefore: P = A * Ar * Ar^2 *...*Ar^8 Collect all the A's P = (A^9)*(1 * r * r^2 ...* r^8) P = A^9 * r^(0+1+2+...+8) There's a formula for the sum of the first n integers (n/2)(n+1), or if you don't know just add it up. 1+2+...+8 = 36 Therefore P = A^9 * r^36 Since 36 is a multiple of 9, you can simplify: P = (Ar^4)^9 Still with me? Remember that Ar^4=2 (a given fact). So finally P = 2^9 = 512. Cute problem.


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iF THE QUESTION IS WRITTEN LIKE THIS: WHAT IS THE VALUE IN r IN THE INEQUALITY 5>r=3. THEN THE BEST POSSIBLE ANSWER WOULD BE...D) R<8


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Is r equals 9 a solution of 128 equals 16r?

No, it is not. Substitute and check. You have 128 + 16r. So now, if r=9, then the equation would be 128 = 16x9. But, 16x9 is 145 and is not equal to 128. Thus, r=9 is not the solution. The solution is r+ 128/16 which is equal to 8. Thus, the actual solution is r = 8.