The nth term of a sequence is $2n^2 - 1$ - Edexcel - GCSE Maths - Question 7 - 2019 - Paper 2
Question 7
The nth term of a sequence is $2n^2 - 1$.
The nth term of a different sequence is $40 - n^2$.
Show that there is only one number that is in both of these sequences.
Worked Solution & Example Answer:The nth term of a sequence is $2n^2 - 1$ - Edexcel - GCSE Maths - Question 7 - 2019 - Paper 2
Step 1
Finding common terms in the first sequence
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Answer
The nth term of the first sequence can be expressed as:
an=2n2−1
To find the first few terms, we can substitute values of n:
For n=1: 2(12)−1=1
For n=2: 2(22)−1=7
For n=3: 2(32)−1=17
For n=4: 2(42)−1=31
For n=5: 2(52)−1=49
For n=6: 2(62)−1=71
For n=7: 2(72)−1=97
For n=8: 2(82)−1=127
For n=9: 2(92)−1=161
For n=10: 2(102)−1=199
Step 2
Finding common terms in the second sequence
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Answer
The nth term of the second sequence is:
bn=40−n2
Calculating the first few terms:
For n=1: 40−(12)=39
For n=2: 40−(22)=36
For n=3: 40−(32)=31
For n=4: 40−(42)=24
For n=5: 40−(52)=15
For n=6: 40−(62)=4
For n=7: 40−(72)=−9
For n=8: 40−(82)=−24
For n=9: 40−(92)=−41
For n=10: 40−(102)=−60
Step 3
Finding the common number
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