Question

If the sum of the first m terms of an AP is \[2{{m}^{2}}+3m,\] then what is its second term?

Answer 1

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.