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.
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let the first term be a , and the common ratio r .
then we have a r 4 = 1 0 .
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 3 6 = ( a r 4 ) 9 = 1 0 9
The n th term of a geometric seqence is given by,
n th term = 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 ......... (1)
Now, the first 9 terms of the sequence would be a, a r 1 , a r 2 , ...., a r 8 .
Thus, their product would be a 9 r 3 6 which can be rewritten as, ( a r 4 ) 9 .
Now in the above rearrangement, we can substitute the value from (1) and thus we obtain 1 0 9 = 1 0 0 0 0 0 0 0 0 0
Yay! Great job there! Can you generalize this problem too?
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In what sense, should I generalise, or what do you exactly want to me generalise?
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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 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 . 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 so their product would be a M r 2 M ( M − 1 ) 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 2 M − 1 ) M . Now we already know a r N − 1 so if a r N − 1 and a r 2 M − 1 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.
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Yup, this is great! I'm glad you enjoyed this question (and the generalized question).
Let the first term and the common ratio of the geometric progression be a and r respectively. Then the fifth term is a r 4 = 1 0 . 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 2 8 ( 8 + 1 ) = a 9 r 3 6 = ( a r 4 ) 9 = 1 0 9 = 1 0 0 0 0 0 0 0 0 0
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 =10. The answer then is 1 0 4 =10000
Will the answer be different if the first term is not equal to 1?
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Please mention in question that common difference can not be 1
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?
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Yeah, this question is still wrong. The staffs didn't fix the answer.
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I think you have changed the question. It was different before
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If the fifth term is 1 0 , then the sixth term is 1 0 r , the seventh term is 1 0 r 2 , the eighth term is 1 0 r 3 , and the ninth term is 1 0 r 4 ; and likewise the fourth term is r 1 0 , the third term is r 2 1 0 , the second term is r 3 1 0 , and the first term is r 4 1 0 .
The product of the first 9 terms is then:
r 4 1 0 ⋅ r 3 1 0 ⋅ r 2 1 0 ⋅ r 1 0 ⋅ 1 0 ⋅ 1 0 r ⋅ 1 0 r 2 ⋅ 1 0 r 3 ⋅ 1 0 r 4
All the r terms cancel out, leaving 1 0 9 = 1 0 0 0 0 0 0 0 0 0 .