
If the sum of the first m terms of an AP is \[2{{m}^{2}}+3m,\] then what is its second term?
Answer
483.6k+ views
Hint: We are given the sum of m terms denoted by \[2{{m}^{2}}+3m,\] so we will put m = 1 and will get the sum of the first term which is nothing but the first term only. Then, we put m = 2 and get the sum of the first two terms of the AP. Finally, we will subtract the first term from the sum of the two terms to get the second term.
Complete step-by-step answer:
We are given that the sum of the first m term of the AP is given as \[2{{m}^{2}}+3m.\] We are asked to find the second term to do so we need the common difference and the first term.
Now, \[2{{m}^{2}}+3m\] denotes the sum of the first m terms. When we put m = 1, it means we get the sum of the first term only. So,
\[{{S}_{1}}=2\times {{1}^{2}}+3\times 1\]
\[\Rightarrow {{S}_{1}}=2+3\]
\[\Rightarrow {{S}_{1}}=5\]
So, the sum of the first term is 5. This means we have got \[{{a}_{1}}=5.\left[ {{a}_{1}}=\text{first term} \right]\]
Now, when we put m = 2 in \[2{{m}^{2}}+3m,\] we will get the sum of the first two terms of the AP. Therefore, we get,
\[{{S}_{2}}=2{{m}^{2}}+3m\]
\[\Rightarrow {{S}_{2}}=2\times {{2}^{2}}+3\times 2\]
Solving further, we get,
\[\Rightarrow {{S}_{2}}=2\times 4+6\]
\[\Rightarrow {{S}_{2}}=14\]
So, we get the sum of the first two terms as 14.
We will take \[{{a}_{1}}\] as the first term and \[{{a}_{2}}\] as the second term. Then we have,
\[{{a}_{1}}+{{a}_{2}}={{S}_{2}}\]
As \[{{S}_{2}}=14,{{a}_{1}}=5,\] we get,
\[\Rightarrow {{a}_{2}}=14-5\]
\[\Rightarrow {{a}_{2}}=9\]
Hence, we get that the second term of the AP is 9.
So, the required answer is 9.
Note: We can also find the second term by first finding ‘a’ (first term) and ‘d’ (common difference).
When we put m = 1, we get,
\[{{S}_{1}}=2{{\times }^{2}}+3\times 1\]
\[\Rightarrow {{S}_{1}}=5\]
So, our first term is 5. Therefore, a is 5.
Then in an AP, the second term is denoted by d. When we put m = 2, we get,
\[{{S}_{2}}=2\times {{2}^{2}}+3\times 2\]
\[\Rightarrow {{S}_{2}}=2\times 4+6\]
\[\Rightarrow {{S}_{2}}=14\]
\[{{S}_{2}}=\text{Sum of first term + second term}\]
\[\Rightarrow a+\left( a+d \right)=14\]
\[\Rightarrow 5+5+d=14\]
\[\Rightarrow 10+d=14\]
\[\Rightarrow d=14-10\]
\[\Rightarrow d=4\]
Now, we have a = 5 and d = 4. So, we can easily find the second term,
\[{{a}_{2}}=a+d\]
\[\Rightarrow {{a}_{2}}=5+4\]
\[\Rightarrow {{a}_{2}}=9\]
So, the second term is 9.
Complete step-by-step answer:
We are given that the sum of the first m term of the AP is given as \[2{{m}^{2}}+3m.\] We are asked to find the second term to do so we need the common difference and the first term.
Now, \[2{{m}^{2}}+3m\] denotes the sum of the first m terms. When we put m = 1, it means we get the sum of the first term only. So,
\[{{S}_{1}}=2\times {{1}^{2}}+3\times 1\]
\[\Rightarrow {{S}_{1}}=2+3\]
\[\Rightarrow {{S}_{1}}=5\]
So, the sum of the first term is 5. This means we have got \[{{a}_{1}}=5.\left[ {{a}_{1}}=\text{first term} \right]\]
Now, when we put m = 2 in \[2{{m}^{2}}+3m,\] we will get the sum of the first two terms of the AP. Therefore, we get,
\[{{S}_{2}}=2{{m}^{2}}+3m\]
\[\Rightarrow {{S}_{2}}=2\times {{2}^{2}}+3\times 2\]
Solving further, we get,
\[\Rightarrow {{S}_{2}}=2\times 4+6\]
\[\Rightarrow {{S}_{2}}=14\]
So, we get the sum of the first two terms as 14.
We will take \[{{a}_{1}}\] as the first term and \[{{a}_{2}}\] as the second term. Then we have,
\[{{a}_{1}}+{{a}_{2}}={{S}_{2}}\]
As \[{{S}_{2}}=14,{{a}_{1}}=5,\] we get,
\[\Rightarrow {{a}_{2}}=14-5\]
\[\Rightarrow {{a}_{2}}=9\]
Hence, we get that the second term of the AP is 9.
So, the required answer is 9.
Note: We can also find the second term by first finding ‘a’ (first term) and ‘d’ (common difference).
When we put m = 1, we get,
\[{{S}_{1}}=2{{\times }^{2}}+3\times 1\]
\[\Rightarrow {{S}_{1}}=5\]
So, our first term is 5. Therefore, a is 5.
Then in an AP, the second term is denoted by d. When we put m = 2, we get,
\[{{S}_{2}}=2\times {{2}^{2}}+3\times 2\]
\[\Rightarrow {{S}_{2}}=2\times 4+6\]
\[\Rightarrow {{S}_{2}}=14\]
\[{{S}_{2}}=\text{Sum of first term + second term}\]
\[\Rightarrow a+\left( a+d \right)=14\]
\[\Rightarrow 5+5+d=14\]
\[\Rightarrow 10+d=14\]
\[\Rightarrow d=14-10\]
\[\Rightarrow d=4\]
Now, we have a = 5 and d = 4. So, we can easily find the second term,
\[{{a}_{2}}=a+d\]
\[\Rightarrow {{a}_{2}}=5+4\]
\[\Rightarrow {{a}_{2}}=9\]
So, the second term is 9.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Why is there a time difference of about 5 hours between class 10 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Write an application to the principal requesting five class 10 english CBSE
