Question

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

Answer 1

Hint: In order to find the second term, we will substitute the value of \[m\]by one and two. Then we get the first term and the sum of first and second terms respectively. Next by subtracting the first term from the sum we get the second term.

Complete step by step answer:
Let us consider the sequence as ${t_1},{t_2},{t_{3,...}}$
As an initial step, we will find the first term of the series by using the given sum. Then we will proceed.
It is given that, the sum of \[m\]terms of AP are,
\[{S_m} = 2{m^2} + 3m\]
Substitute \[m = 1\] in the given sum to find the first term we get,
\[{S_1} = 2{(1)^2} + 3(1) = 5\]
Here \[{S_1}\]is nothing but the first term of the sequence${t_1}$.
So, the first term of the sequence is \[{t_1} = 5\].
To find the second term we should find the sum of two terms, so
Again, substitute \[m = 2\]in the given sum we get,
\[{S_2} = 2{(2)^2} + 3(2) = 14\]
It implies the sum of first term and second term is \[14.\]
So,
\[{t_1} + {t_2} = 14\]
Initially we have found that \[{t_1} = 5\]
Substitute the value of first term that is \[{t_1}\] in the above equation we get,
${t_2} = 14 - {t_1}$
\[{t_2} = 14 - 5 = 9\]
Hence we have found that the second term \[{t_2}\] is 9.
Hence,
The first term and second term of the given AP series are \[5\] and \[9\] respectively.

Note: The general form of arithmetic progression is,
\[a,(a + d),(a + 2d),...[a + (n - 1)d]\], where, \[a,d\] are known as the initial term and common difference.
Then the sum of nth terms is given by the formula,
\[{S_n} = \dfrac{n}{2}[2a + (n - 1)d]\].