Question

If the sum of p terms of an AP is \[3{{p}^{2}}+4p\]. Find its \[{{n}^{th}}\] term.

Answer 1

Hint: Write the \[{{n}^{th}}\] term of a sequence by using the sum of n terms of the same sequence. Here you know the formula for the sum of p terms. Use this expression to find the sum of n terms. Form the first relation, find the expression for the \[{{n}^{th}}\] term. This expression is our required result.

Complete step-by-step answer:
Sum of n terms is defined as the sum of all the terms from the first term up till \[{{n}^{th}}\] term. Similarly, the sum of (n – 1) terms is defined as the sum of all terms from the first term up till \[{{\left( n-1 \right)}^{th}}\] term.
Writing the above statement mathematically with \[{{S}_{n}}\] as sum of n terms, \[{{S}_{n-1}}\] as the sum of (n – 1) terms, \[{{a}_{k}}\] as \[{{k}^{th}}\] term, we get the equations as:
\[{{S}_{n}}={{a}_{1}}+{{a}_{2}}+.....+{{a}_{n-1}}+{{a}_{n}}.....\left( i \right)\]
\[{{S}_{n-1}}={{a}_{1}}+{{a}_{2}}+.....+{{a}_{n-1}}.....\left( ii \right)\]
As you can see the only difference between (i) and (ii) is the \[{{n}^{th}}\] term of the sequence given. By using this, find the \[{{n}^{th}}\] term. By subtracting equation (ii) from (i), we get it as:
\[{{S}_{n}}-{{S}_{n-1}}=\left( {{a}_{1}}+{{a}_{2}}+....+{{a}_{n}} \right)-\left( {{a}_{1}}+{{a}_{2}}+.....+{{a}_{n-1}} \right)\]
By canceling the common terms in the above equation, we get,
\[{{S}_{n}}-{{S}_{n-1}}={{a}_{n}}....\left( iii \right)\]
By this, we can write the statement given by (for the sequence):
\[{{n}^{th}}\text{ term}=\left( \text{Sum of n terms} \right)-\left( \text{Sum of }\left( n-1 \right)\text{ terms} \right)\]
The given expression for the sum of p terms of an AP in the question is:
\[{{S}_{p}}=3{{p}^{2}}+4p\]
From this, we get the values of the sum of n terms, (n – 1) terms as:
\[{{S}_{n}}=3{{n}^{2}}+4n;{{S}_{n-1}}=3{{\left( n-1 \right)}^{2}}+4\left( n-1 \right)\]
By substituting these into equation (iii), we get it as:
\[{{a}_{n}}=3{{n}^{2}}+4n-3{{\left( n-1 \right)}^{2}}-4\left( n-1 \right)\]
By using the algebraic identity given by \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\].
\[{{a}_{n}}=3{{n}^{2}}+4n-3\left( {{n}^{2}}-2n+1 \right)-4\left( n-1 \right)\]
By removing the bracket of the term with 3, we get the equation:
\[{{a}_{n}}=3{{n}^{2}}+4n-3{{n}^{2}}+6n-3-4\left( n-1 \right)\]
By removing the last bracket, we get the equation as:
\[{{a}_{n}}=3{{n}^{2}}+4n-3{{n}^{2}}+6n-3-4n+4\]
By bringing similar terms together, we get the equation as:
\[{{a}_{n}}=\left( 3{{n}^{2}}-3{{n}^{2}} \right)+\left( 4n+6n-4n \right)+\left( 4-3 \right)\]
By taking the terms of n as common, we can write:
\[{{a}_{n}}=\left( 3-3 \right){{n}^{2}}+\left( 4+6-4 \right)n+\left( 4-3 \right)\]
By simplifying the above equation, we can write it as,
\[{{a}_{n}}=6n+1\]
Therefore, the \[{{n}^{th}}\] term is written as 6n + 1.

Note: The idea of subtracting both the sums to get the \[{{n}^{th}}\] term is very important. Whenever you have two equations, you must always see the point of difference to get these ideas. Be careful while expanding and multiplying 3 with all the terms inside the bracket, do not forget to multiply it with constant as well or else you will get the wrong result.