Question

Which term of the AP 4, 12, 20, 28, … will be 120 more than its \[{21^{st}}\] term?

Answer 1

Hint: We can find the \[{1^{st}}\] term a and common difference d from given AP. Then we can find the \[{21^{st}}\] term using the formula to find the nth term of an AP. Then we can add 120 to it and equate to the equation of nth term. Then we can solve for n to find the required term.

Complete step-by-step answer:
We are given the AP 4, 12, 20, 28, …
Here the \[{1^{st}}\] term is $a = 4$.
We know that the common difference is given by the difference between two consecutive terms.
$ \Rightarrow d = {a_3} - {a_2}$
On substituting the \[{2^{nd}}\] and \[{3^{rd}}\] terms, we get,
$ \Rightarrow d = 20 - 12$
On simplification we get,
$ \Rightarrow d = 8$.
We know that the nth term of an AP is given by the equation ${a_n} = a + \left( {n - 1} \right)d$
Therefore, the \[{21^{st}}\] term is given by,
$ \Rightarrow {a_{21}} = a + \left( {21 - 1} \right)d$
On substituting the values of a and d. we get,
$ \Rightarrow {a_{21}} = 4 + \left( {20} \right) \times 8$
On simplification, we get
$ \Rightarrow {a_{21}} = 4 + 160$
On adding we get,
$ \Rightarrow {a_{21}} = 164$
We need to find the term which is 120 more than the \[{21^{st}}\] term. Let the required term be the ${m^{th}}$ term. Then we can write,
${a_m} = {a_{21}} + 120$
We know that the ${n^{th}}$ term of an AP is given by the equation ${a_n} = a + \left( {n - 1} \right)d$
$ \Rightarrow a + \left( {m - 1} \right)d = {a_{21}} + 120$
On substituting the values, we get
$ \Rightarrow 4 + \left( {m - 1} \right) \times 8 = 164 + 120$
On expanding the bracket, we get
$ \Rightarrow 4 + 8m - 8 = 164 + 120$
On taking the constant to one side, we get
$ \Rightarrow 8m = 164 + 120 + 8 - 4$
On simplification, we get
\[ \Rightarrow 8m = 288\]
On dividing throughout with 8, we get
\[ \Rightarrow m = \dfrac{{288}}{8}\]
On division, we get
\[ \Rightarrow m = 36\]
Now by substituting the value of m in the equation of general term, we get
$ \Rightarrow {a_{36}} = a + \left( {36 - 1} \right)d$
On substituting the values of a and d, we get
$ \Rightarrow {a_{36}} = 4 + \left( {35} \right) \times 8$
On simplification, we get
$ \Rightarrow {a_{36}} = 4 + 280$
On adding, we get
$ \Rightarrow {a_{36}} = 284$
Therefore, the required term is \[{36^{th}}\] term which is equal to 284.

Note: Alternate method to solve this problem is given by,
We are given the AP 4, 12, 20, 28, …
Here the \[{1^{st}}\] term is $a = 4$.
We know that the common difference is given by the difference between two consecutive terms.
$ \Rightarrow d = {a_3} - {a_2}$
On substituting the \[{2^{nd}}\] and \[{3^{rd}}\] terms, we get
$ \Rightarrow d = 20 - 12$
On simplification, we get
$ \Rightarrow d = 8$
We need to find the term which is 120 more than the \[{21^{st}}\] term. Let the required term be the ${m^{th}}$
 term. Then we can write,
${a_m} = {a_{21}} + 120$
We know that the nth term of an AP is given by the equation ${a_n} = a + \left( {n - 1} \right)d$
$ \Rightarrow a + \left( {m - 1} \right)d = a + \left( {20} \right)d + 120$
On rearranging, we get
$ \Rightarrow \left( {m - 1} \right)d - 20d = 120$
On simplification, we get
$ \Rightarrow \left( {m - 1 - 20} \right)d = 120$
On substituting the value of d, we get
$ \Rightarrow \left( {m - 21} \right) \times 8 = 120$
On dividing throughout by 8, we get
$ \Rightarrow \left( {m - 21} \right) = \dfrac{{120}}{8}$
On simplification, we get
$ \Rightarrow \left( {m - 21} \right) = 15$
On adding 21 to both sides, we get
$ \Rightarrow m = 15 + 21$
On simplification, we have
$ \Rightarrow m = 36$
Now by substituting the value of m in the equation of general term, we get
$ \Rightarrow {a_{36}} = a + \left( {36 - 1} \right)d$
On substituting the values of a and d, we get
$ \Rightarrow {a_{36}} = 4 + \left( {35} \right) \times 8$
On simplification, we get
$ \Rightarrow {a_{36}} = 4 + 280$
On addition, we have
$ \Rightarrow {a_{36}} = 284$
Therefore, the required term is \[{36^{th}}\] term which is equal to 284.