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All the terms of a geometric series are positive - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 2
Question 7
All the terms of a geometric series are positive. The sum of the first two terms is 34 and the sum to infinity is 162. Find a) the common ratio, b) the first term... show full transcript
Worked Solution & Example Answer:All the terms of a geometric series are positive - Edexcel - A-Level Maths Pure - Question 7 - 2015 - Paper 2
Step 1
Find the common ratio
Answer
Let the first term be ( a ) and the common ratio be ( r ).
From the problem, we have the following equations:
- The sum of the first two terms:
[ a + ar = 34 ] - The sum to infinity:
[ \frac{a}{1 - r} = 162 ]
From the second equation, we can express ( a ):
[ a = 162(1 - r) ]
Now, substitute this into the first equation:
[ 162(1 - r) + 162(1 - r)r = 34 ]
[ 162(1 - r)(1 + r) = 34 ]
[ 162(1 - r^2) = 34 ]
[ 1 - r^2 = \frac{34}{162} ]
This simplifies to:
[ r^2 = 1 - \frac{17}{81} = \frac{64}{81} ]
[ r = \frac{8}{9} ]
Thus, the common ratio is ( r = \frac{8}{9} ).
Step 2
Step 3
Find the smallest value of n where sum exceeds 290
Answer
For the second geometric series with first term 42 and common ratio ( \frac{6}{7} ):
The sum of the first ( n ) terms ( S_n ) is given by:
[ S_n = a \frac{1 - r^n}{1 - r} ]
[ S_n = 42 \frac{1 - (\frac{6}{7})^n}{1 - \frac{6}{7}} = 42 \cdot 7(1 - (\frac{6}{7})^n) = 294(1 - (\frac{6}{7})^n) ]
We need to find the smallest ( n ) where ( S_n > 290 ):
[ 294(1 - (\frac{6}{7})^n) > 290 ]
[ 1 - (\frac{6}{7})^n > \frac{290}{294} \approx 0.986 ]
[ (\frac{6}{7})^n < 0.014 ]
Taking logarithm on both sides:
[ n \log(\frac{6}{7}) < \log(0.014) ]
Since ( \log(\frac{6}{7}) ) is negative, we reverse the inequality:
[ n > \frac{\log(0.014)}{\log(\frac{6}{7})} \approx 28 ]
Thus, the smallest value of ( n ) is ( n = 28 ).