How do you write the expression for the $n^{th}$ term of the sequence given 2, -4, 6, -8, 10… ?
Answer
610.8k+ views
Hint: We will try to form a relation between the first few given terms such that when we are writing the terms in sequence we either form a sequence in increasing order or in decreasing order. Here we observe the common number between all the terms is 2 and one is negative one is positive, so we try to form terms with multiplication to 2 and having powers of -1.
Complete step-by-step answer:
We are given the sequence 2, -4, 6, -8, 10…
We will try to write each of the first few given terms in a pattern that is similar for all terms.
First term:
We are given the first term as 2
We can write \[2 = {( - 1)^{(1) + 1}}\left[ {2 \times (1)} \right]\]
Second term:
We are given the second term as -4
We can write \[ - 4 = {( - 1)^{(2) + 1}}\left[ {2 \times (2)} \right]\]
Third term:
We are given the third term as 6
We can write \[6 = {( - 1)^{(3) + 1}}\left[ {2 \times (3)} \right]\]
Fourth term:
We are given the fourth term as -8
We can write \[ - 8 = {( - 1)^{(4) + 1}}\left[ {2 \times (4)} \right]\]
Let us denote the first, second, third … terms of the sequence by \[{a_1},{a_2},{a_3}...\]
Then \[{a_1} = {( - 1)^{(1) + 1}}\left[ {2 \times (1)} \right];{a_2} = {( - 1)^{(2) + 1}}\left[ {2 \times (2)} \right];{a_3} = {( - 1)^{(3) + 1}}\left[ {2 \times (3)} \right]...\]
Then we can write the nth term of this sequence i.e. say \[{a_n}\]as
\[{a_n} = {( - 1)^{(n) + 1}}\left[ {2 \times (n)} \right]\]
\[\therefore \]The nth term of the sequence 2, -4, 6, -8, 10… can be given by the expression\[{a_n} = {( - 1)^{(n) + 1}}\left[ {2 \times (n)} \right]\], where \[{a_n}\] is the nth term of the sequence.
Note:
Many students make the mistake of assuming the sequence given as an arithmetic, geometric or harmonic progression and they try to find the value of d from the first two terms and then directly write the nth term using the formula which is wrong. Keep in mind we should always check all given pairs of terms in order to state and prove that sequence is an A.P, G.P or an H.P.
Complete step-by-step answer:
We are given the sequence 2, -4, 6, -8, 10…
We will try to write each of the first few given terms in a pattern that is similar for all terms.
First term:
We are given the first term as 2
We can write \[2 = {( - 1)^{(1) + 1}}\left[ {2 \times (1)} \right]\]
Second term:
We are given the second term as -4
We can write \[ - 4 = {( - 1)^{(2) + 1}}\left[ {2 \times (2)} \right]\]
Third term:
We are given the third term as 6
We can write \[6 = {( - 1)^{(3) + 1}}\left[ {2 \times (3)} \right]\]
Fourth term:
We are given the fourth term as -8
We can write \[ - 8 = {( - 1)^{(4) + 1}}\left[ {2 \times (4)} \right]\]
Let us denote the first, second, third … terms of the sequence by \[{a_1},{a_2},{a_3}...\]
Then \[{a_1} = {( - 1)^{(1) + 1}}\left[ {2 \times (1)} \right];{a_2} = {( - 1)^{(2) + 1}}\left[ {2 \times (2)} \right];{a_3} = {( - 1)^{(3) + 1}}\left[ {2 \times (3)} \right]...\]
Then we can write the nth term of this sequence i.e. say \[{a_n}\]as
\[{a_n} = {( - 1)^{(n) + 1}}\left[ {2 \times (n)} \right]\]
\[\therefore \]The nth term of the sequence 2, -4, 6, -8, 10… can be given by the expression\[{a_n} = {( - 1)^{(n) + 1}}\left[ {2 \times (n)} \right]\], where \[{a_n}\] is the nth term of the sequence.
Note:
Many students make the mistake of assuming the sequence given as an arithmetic, geometric or harmonic progression and they try to find the value of d from the first two terms and then directly write the nth term using the formula which is wrong. Keep in mind we should always check all given pairs of terms in order to state and prove that sequence is an A.P, G.P or an H.P.
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