Find the nth term of the G.P. 1, 2, 4, 8, 16…………
Answer
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Hint: This is a straightforward question just use the formula of general term of a G.P. which is ${{T}_{n}}=a{{r}^{n-1}}$ where “a” is the first term of the G.P. and “r” is the common ratio of the G.P.
Complete step-by-step answer:
In the given G.P., the first term is 1 and the common ratio is calculated as dividing any term of the sequence by the preceding term.
1, 2, 4, 8, 16……………….
From the above sequence, we can find the common ratio by taking any term say 2 and dividing its preceding term which is 1 which will give a common ratio as 2. Similarly, you can check the common ratio for the other terms of the sequence too like take 8 and divide by its preceding term which is 4 then the common ratio which will get is 2.
The general term of the G.P. is ${{T}_{n}}=a{{r}^{n-1}}$. Here a = 1 and r = 2 as we have discussed above. So, substituting the “a” and “r” values in general term expression we get:
$\begin{align}
& {{T}_{n}}=\left( 1 \right){{\left( 2 \right)}^{n-1}} \\
& \Rightarrow {{\text{T}}_{n}}={{2}^{n-1}} \\
\end{align}$
Hence, the general term of the given G.P. is 2n-1.
Note: We can verify whether the general term that we have solved is correct by putting different values of n in the general term 2n-1 which leads to a sequence of numbers and then we will match with the one given in question.
Now, what values can n have? n can take values 1, 2, 3, 4,……. Because n represents the order of the sequence like putting n=1 in the general term gives the first term, n=2 will give the second term and so on.
The sequence that we are getting from putting different values of n as follows:
n = 1, T1 = 20 = 1
n = 2, T2 = 21 = 2
n = 3, T3 = 22 = 4
n = 4, T4 = 23 = 8
So the sequence we are obtaining is 1, 2, 4, 8…… which is matching with the given sequence. Hence, the general term that we have solved is correct.
Complete step-by-step answer:
In the given G.P., the first term is 1 and the common ratio is calculated as dividing any term of the sequence by the preceding term.
1, 2, 4, 8, 16……………….
From the above sequence, we can find the common ratio by taking any term say 2 and dividing its preceding term which is 1 which will give a common ratio as 2. Similarly, you can check the common ratio for the other terms of the sequence too like take 8 and divide by its preceding term which is 4 then the common ratio which will get is 2.
The general term of the G.P. is ${{T}_{n}}=a{{r}^{n-1}}$. Here a = 1 and r = 2 as we have discussed above. So, substituting the “a” and “r” values in general term expression we get:
$\begin{align}
& {{T}_{n}}=\left( 1 \right){{\left( 2 \right)}^{n-1}} \\
& \Rightarrow {{\text{T}}_{n}}={{2}^{n-1}} \\
\end{align}$
Hence, the general term of the given G.P. is 2n-1.
Note: We can verify whether the general term that we have solved is correct by putting different values of n in the general term 2n-1 which leads to a sequence of numbers and then we will match with the one given in question.
Now, what values can n have? n can take values 1, 2, 3, 4,……. Because n represents the order of the sequence like putting n=1 in the general term gives the first term, n=2 will give the second term and so on.
The sequence that we are getting from putting different values of n as follows:
n = 1, T1 = 20 = 1
n = 2, T2 = 21 = 2
n = 3, T3 = 22 = 4
n = 4, T4 = 23 = 8
So the sequence we are obtaining is 1, 2, 4, 8…… which is matching with the given sequence. Hence, the general term that we have solved is correct.
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