
Find the nth term for the sequence: \[3,8,15,{\text{ }}24...\]
Answer
455.1k+ views
Hint:Check for the difference between the terms that are known, verify if it is an arithmetic progression. Check if they are related by a common factor, to verify if it is a geometric progression. Check for both, to verify if it is a mixture of arithmetic-geometric progression. In case the difference of the terms of the sequence are also a sequence, identify the type and proceed accordingly.
Complete step by step solution:
Let us first subtract the terms from its preceding term and verify if there is a common difference.
$8 - 3 = 5$
$15 - 8 = 7$
$24 - 15 = 9$
And so on...
Thus, we see that the series doesn’t have a common difference. Rather the differences also form another sequence, that is $5,7,9..$.
In such cases we cannot apply the formula for the nth term of an arithmetic progression directly.
Let the series in the question have the terms named as ${T_1},{T_2},{T_3},{T_4}$..respectively.
Now, ${T_1}$ can be written as,
${T_1} = {2^2} - 1 = 3$
Similarly,
$
{T_2} = {3^2} - 1 = 8 \\
{T_3} = {4^2} - 1 = 15 \\
{T_4} = {5^2} - 1 = 24 \\
. \\
. \\
. \\
{T_n} = {(n + 1)^2} - 1 \\
$
Thus, ${T_n} = {n^2} + 2n + 1 - 1 = {n^2} + 2n$, where ${T_n}$is the nth term of the sequence.
Alternate method: Let us consider the sum of the given sequence as, S.
\[{S_1} = 3 + 8 + 15 + 24 + \ldots .. + {T_{n - 1}} + {T_n}\]….Consider this equation (i).
Again, \[{S_2} = 0 + 3 + 8 + 15 + 24 + \ldots .. + {T_{n - 1}} + {T_n}\]……….equation (ii)
Subtracting (ii) from (i), we get
\[{S_1} - {S_2} = 0 = 3 + 5 + 7 + 9 + \ldots .. - {T_n}\]
We see that ${T_n}$ is the only term that is negative on the RHS. Taking it to the LHS, we have
${T_n} = 3 + 5 + 7 + 9...$
Now, ${T_n}$ is the sum of odd natural numbers that is equal to ${n^2} + 2n = n(n + 2)$
Note: The “sum of n terms of AP” equals to the sum(addition) of first n terms of the arithmetic sequence. Its adequate n divided by 2 times the sum of twice the primary term, 'a' and therefore the product of the difference between second and first term-'d' also called common difference, and (n-1), where n is the number of terms to be added.
Complete step by step solution:
Let us first subtract the terms from its preceding term and verify if there is a common difference.
$8 - 3 = 5$
$15 - 8 = 7$
$24 - 15 = 9$
And so on...
Thus, we see that the series doesn’t have a common difference. Rather the differences also form another sequence, that is $5,7,9..$.
In such cases we cannot apply the formula for the nth term of an arithmetic progression directly.
Let the series in the question have the terms named as ${T_1},{T_2},{T_3},{T_4}$..respectively.
Now, ${T_1}$ can be written as,
${T_1} = {2^2} - 1 = 3$
Similarly,
$
{T_2} = {3^2} - 1 = 8 \\
{T_3} = {4^2} - 1 = 15 \\
{T_4} = {5^2} - 1 = 24 \\
. \\
. \\
. \\
{T_n} = {(n + 1)^2} - 1 \\
$
Thus, ${T_n} = {n^2} + 2n + 1 - 1 = {n^2} + 2n$, where ${T_n}$is the nth term of the sequence.
Alternate method: Let us consider the sum of the given sequence as, S.
\[{S_1} = 3 + 8 + 15 + 24 + \ldots .. + {T_{n - 1}} + {T_n}\]….Consider this equation (i).
Again, \[{S_2} = 0 + 3 + 8 + 15 + 24 + \ldots .. + {T_{n - 1}} + {T_n}\]……….equation (ii)
Subtracting (ii) from (i), we get
\[{S_1} - {S_2} = 0 = 3 + 5 + 7 + 9 + \ldots .. - {T_n}\]
We see that ${T_n}$ is the only term that is negative on the RHS. Taking it to the LHS, we have
${T_n} = 3 + 5 + 7 + 9...$
Now, ${T_n}$ is the sum of odd natural numbers that is equal to ${n^2} + 2n = n(n + 2)$
Note: The “sum of n terms of AP” equals to the sum(addition) of first n terms of the arithmetic sequence. Its adequate n divided by 2 times the sum of twice the primary term, 'a' and therefore the product of the difference between second and first term-'d' also called common difference, and (n-1), where n is the number of terms to be added.
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