
What are the next three terms in the sequence $ 1.0,2.5,4.0,5.5? $
Answer
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Hint: As we can see that the above given sequence is in A.P. We know that an arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. Here in this question we have to find the next three terms, also some common terms of A.P is - $ n $ is the number of terms, $ a = $ first term and $ d $ is the common difference. We will first find the common difference between the two terms of the given series and then we can apply the formula $ {a_n} = a + (n - 1)d $ .
Complete step-by-step answer:
As the per given question we have the sequence: $ 1.0,2.5,4.0,5.5... $ Here we have to find the three terms so $ n $ will be $ 5,6\,\,and\,\,7 $ ., also we know the formula of $ nth $ term is $ a + (n - 1)d $ .
We have $ a = 1,d = 2.5 - 1.0 = 1.5 $ and $ n = 5,6,7 $ .
So we will first find the fifth term i.e. $ {a_5} $ . Now by applying the formula we have
$ {a_5} = 1 + (5 - 1)1.5 $ .
Now we solve it, it gives us
$ {a_5} = 1 + 4 \times 1.5 \Rightarrow {a_5} = 1 + 6 $ .
So the fifth term is $ 1 + 6 = 7 $ .
Similarly we find the sixth term i.e. $ {a_6} $ . Now by applying the formula we have
$ {a_6} = 1 + (6 - 1)1.5 $ .
Now we solve it, it gives us
$ {a_6} = 1 + 5 \times 1.5 \Rightarrow {a_6} = 1 + 7.5 $ .
So the sixth term is $ 1 + 7.5 = 8.5 $ .
Again for the seventh term i.e. $ {a_7} $ . Now by applying the formula we have
$ {a_7} = 1 + (7 - 1)1.5 $ .
Now we solve it, it gives us
${a_7} = 1 + 6 \times 1.5 \Rightarrow {a_7} = 1 + 9 $ .
So the seventh term is $ 1 + 9 = 10 $ .
Hence the next three terms of the given A.P is $ 7.0,8.5,10.0 $ .
So, the correct answer is “ $ 7.0,8.5,10.0 $ ”.
Note: We should be aware of the arithmetic sequence and their formula before solving this kind of question. We should carefully substitute the values and solve them. We should know that we can find the sum of the given arithmetic sequence by adding the first and last term and then divide the sum by two, this is also a formula when the first and the last term is given in the question. And then the sum of the sequence will be the number of terms multiplied by the average number of terms in the sequence. The formula can be written as $ S = \dfrac{n}{2}(a + l) $ where the number of terms and $ a,l $ are first and last terms.
Complete step-by-step answer:
As the per given question we have the sequence: $ 1.0,2.5,4.0,5.5... $ Here we have to find the three terms so $ n $ will be $ 5,6\,\,and\,\,7 $ ., also we know the formula of $ nth $ term is $ a + (n - 1)d $ .
We have $ a = 1,d = 2.5 - 1.0 = 1.5 $ and $ n = 5,6,7 $ .
So we will first find the fifth term i.e. $ {a_5} $ . Now by applying the formula we have
$ {a_5} = 1 + (5 - 1)1.5 $ .
Now we solve it, it gives us
$ {a_5} = 1 + 4 \times 1.5 \Rightarrow {a_5} = 1 + 6 $ .
So the fifth term is $ 1 + 6 = 7 $ .
Similarly we find the sixth term i.e. $ {a_6} $ . Now by applying the formula we have
$ {a_6} = 1 + (6 - 1)1.5 $ .
Now we solve it, it gives us
$ {a_6} = 1 + 5 \times 1.5 \Rightarrow {a_6} = 1 + 7.5 $ .
So the sixth term is $ 1 + 7.5 = 8.5 $ .
Again for the seventh term i.e. $ {a_7} $ . Now by applying the formula we have
$ {a_7} = 1 + (7 - 1)1.5 $ .
Now we solve it, it gives us
${a_7} = 1 + 6 \times 1.5 \Rightarrow {a_7} = 1 + 9 $ .
So the seventh term is $ 1 + 9 = 10 $ .
Hence the next three terms of the given A.P is $ 7.0,8.5,10.0 $ .
So, the correct answer is “ $ 7.0,8.5,10.0 $ ”.
Note: We should be aware of the arithmetic sequence and their formula before solving this kind of question. We should carefully substitute the values and solve them. We should know that we can find the sum of the given arithmetic sequence by adding the first and last term and then divide the sum by two, this is also a formula when the first and the last term is given in the question. And then the sum of the sequence will be the number of terms multiplied by the average number of terms in the sequence. The formula can be written as $ S = \dfrac{n}{2}(a + l) $ where the number of terms and $ a,l $ are first and last terms.
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