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Question: How do you write the first 5 terms of the geometric sequence where \({{a}_{1}}=5\) , \({{a}_{k+1}}=-...

How do you write the first 5 terms of the geometric sequence where a1=5{{a}_{1}}=5 , ak+1=−2ak{{a}_{k+1}}=-2{{a}_{k}} and determine the common ratio and write the nth{{n}^{th}} term of the sequence as a function of n.

Explanation

Solution

Now we have the first term of the sequence. Now we will use the condition ak+1=−2ak{{a}_{k+1}}=-2{{a}_{k}} to find the ratio of two consecutive terms in the sequence and hence find the common ratio. Now we know that the general term of GP is given by arn−1a{{r}^{n-1}} . Hence we will find the general term by substituting the values in the formula and then find the first 5 terms of the sequence by substituting the values of n.

Complete step by step answer:
Now we are given that the first term of the sequence is 5.
Now we know that the common ratio of GP is nothing but the ratio of two consecutive terms.
Hence r=ak+1akr=\dfrac{{{a}_{k+1}}}{{{a}_{k}}}
Now we are given that ak+1=−2ak{{a}_{k+1}}=-2{{a}_{k}}
⇒ak+1=−2ak ⇒ak+1ak=−2 ⇒r=−2 \begin{aligned} & \Rightarrow {{a}_{k+1}}=-2{{a}_{k}} \\\ & \Rightarrow \dfrac{{{a}_{k+1}}}{{{a}_{k}}}=-2 \\\ & \Rightarrow r=-2 \\\ \end{aligned}
Hence we have r = - 2 and a = 5.
Now we know that the general term of AP is given by,
tn=arn+1{{t}_{n}}=a{{r}^{n+1}}
Hence substituting the values in the formula we get,
tn=5(−2)n−1{{t}_{n}}=5{{\left( -2 \right)}^{n-1}}
Now we know that the first term is 5.
The second term of sequence is 5(−2)=−105\left( -2 \right)=-10
The third term of the sequence is 5(−2)2=205{{\left( -2 \right)}^{2}}=20
The fourth term of the sequence will be 5(−2)3=−405{{\left( -2 \right)}^{3}}=-40
And the fifth term of the sequence will be 5(−2)4=805{{\left( -2 \right)}^{4}}=80 .

Hence the first 5 terms of the sequence are 5,−10,20,−40,805,-10,20,-40,80

Note: Now note that when the ratio of the two consecutive terms is common then we say the sequence is a geometric progression. Similarly if the difference between two consecutive terms is common then we say the sequence is a geometric sequence. General term of arithmetic sequence is given by the formula tn=a+(n−1)d{{t}_{n}}=a+\left( n-1 \right)d and for geometric sequence we have tn=arn+1{{t}_{n}}=a{{r}^{n+1}} .