
Find the nth term for the sequence: \[3,8,15,{\text{ }}24...\]
Answer
447.3k+ 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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
