Why must it be the even term?

Algebra Level 2

Consider a geometric progression with all distinct terms.

Find the product of the first 9 terms of this geometric progression, given that the 5th term of this geometric progression is equal to 10.


The answer is 1000000000.

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5 solutions

David Vreken
Jun 26, 2019

If the fifth term is 10 10 , then the sixth term is 10 r 10r , the seventh term is 10 r 2 10r^2 , the eighth term is 10 r 3 10r^3 , and the ninth term is 10 r 4 10r^4 ; and likewise the fourth term is 10 r \frac{10}{r} , the third term is 10 r 2 \frac{10}{r^2} , the second term is 10 r 3 \frac{10}{r^3} , and the first term is 10 r 4 \frac{10}{r^4} .

The product of the first 9 9 terms is then:

10 r 4 10 r 3 10 r 2 10 r 10 10 r 10 r 2 10 r 3 10 r 4 \frac{10}{r^4} \cdot \frac{10}{r^3} \cdot \frac{10}{r^2} \cdot \frac{10}{r} \cdot 10 \cdot 10r \cdot 10r^2 \cdot 10r^3 \cdot 10r^4

All the r r terms cancel out, leaving 1 0 9 = 1000000000 10^9 = \boxed{1000000000} .

Aareyan Manzoor
Jun 26, 2019

let the first term be a a , and the common ratio r r .

then we have a r 4 = 10 ar^4 =10 .

the product of the first 9 terms is a r × a r 2 × . . . . × a r 8 = a 9 r 1 + 2 + . . . + 8 = a 9 r 36 = ( a r 4 ) 9 = 1 0 9 ar\times ar^2\times .... \times ar^8 = a^9 r^{1+2+...+8}=a^9 r^{36} = \left(a r^4\right)^9 = \boxed{10^9}

The n n th term of a geometric seqence is given by,

n n th term = a r n 1 a{ r }^{ n-1 } (Where a is the first term, and r is the common ratio)

So it is given the 5th term is 10, hence we can rewrite it mathematically using the formula.

10= a r 5 1 = a r 4 a{ r }^{ 5-1 }=a{ r }^{ 4 } ......... (1)

Now, the first 9 terms of the sequence would be a, a r 1 a{ r }^{ 1 } , a r 2 a{ r }^{ 2 } , ...., a r 8 a{ r }^{ 8 } .

Thus, their product would be a 9 r 36 { a }^{ 9 }{ r }^{ 36 } which can be rewritten as, ( a r 4 ) 9 { (a{ r }^{ 4 }) }^{ 9 } .

Now in the above rearrangement, we can substitute the value from (1) and thus we obtain 10 9 { 10 }^{ 9 } = 1000000000 \boxed{1000000000}

Yay! Great job there! Can you generalize this problem too?

Pi Han Goh - 10 months, 4 weeks ago

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In what sense, should I generalise, or what do you exactly want to me generalise?

Siddharth Chakravarty - 10 months, 4 weeks ago

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Find the product of the first M terms of this geometric progression, if you are given the value Nth term of this geometric progression.

If you got M and N, then you can solve the question above. What are all the possibilities of M and N?

Pi Han Goh - 10 months, 4 weeks ago

@Pi Han Goh Here is the generalisation, I have written it shortly, sorry for that! Suppose we know the Nth term of the G.P then basically using the formula of the Nth term of a G.P. we basically know a r N 1 a{ r }^{ N-1 } . We will keep this aside for a moment, and then focus on the next part. We need to find the product of the first M terms, so the first M terms would be a , a r , a r 2 , . . . , a r M 1 a, ar,{ a }{ r }^{ 2 },...,a{ r }^{ M-1 } so their product would be a M r M ( M 1 ) 2 { a }^{ M }r^{ \frac { M(M-1) }2 } as a is multiplied to itself M times and we can find the power of r using the sum of Arithmetic Progression formula. We can take out the common exponent to simplify more as ( a r M 1 2 ) M { (a{ r }^{ \frac { M-1 }{ 2 } }) }^{ M } . Now we already know a r N 1 a{ r }^{ N-1 } so if a r N 1 a{ r }^{ N-1 } and a r M 1 2 { a{ r }^{ \frac { M-1 }{ 2 } } } are equal we can the find the answer, so we can equate both of them to be equal and then cancel a as a is not 0 because all are distinct terms and then r would be left, so the exponents would be same which is basically that, N-1=(M-1)/2 which can be simplified as 2N=M+1, so if M and N follow this relation we can find the answer.

Siddharth Chakravarty - 10 months, 3 weeks ago

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Yup, this is great! I'm glad you enjoyed this question (and the generalized question).

Pi Han Goh - 10 months, 3 weeks ago
Chew-Seong Cheong
Jun 28, 2019

Let the first term and the common ratio of the geometric progression be a a and r r respectively. Then the fifth term is a r 4 = 10 ar^4 = 10 . The product of the first nine terms of the geometric progression is given by:

P = a × a r × a r 2 × a r 3 × × a r 8 = a 9 r 1 + 2 + 3 + + 8 = a 9 r 8 ( 8 + 1 ) 2 = a 9 r 36 = ( a r 4 ) 9 = 1 0 9 = 1000000000 \begin{aligned} P & = a \times ar \times ar^2 \times ar^3 \times \cdots \times ar^8 \\ & = a^9 r^{1+2+3+\cdots + 8} = a^9 r^{\frac {8(8+1)}2} = a^9r^{36} = \left(ar^4\right)^9 = 10^9 = \boxed{1000000000} \end{aligned}

First term of the progression is not given. Assuming this to be 1 and common ratio be r, we have to calculate r^36 given r 9 r^9 =10. The answer then is 1 0 4 10^4 =10000

Will the answer be different if the first term is not equal to 1?

Pi Han Goh - 2 years ago

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Please mention in question that common difference can not be 1

Mr. India - 2 years ago

If the first term be a, then the product of first 9 terms is (a^9)(r^36) and the 10th term is a(r^9) which is 10. Then the product is (a^5)(10^4).

Could you please tell me, what is wrong in this? We have that the 5th therm of this geometric progression is equal to 10 this means a 5=(a 1) r^(4) and we want to calculate product of first 10 therms so it will be (a_1)^(9) r^(36). So the product will be 10^9?

Dan Czinege - 2 years ago

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Yeah, this question is still wrong. The staffs didn't fix the answer.

Pi Han Goh - 2 years ago

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I think you have changed the question. It was different before

Mr. India - 2 years ago

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