Question

In an arithmetic sequence, ${S_n}$ represents the sum to $n$ terms, then the value of ${S_n} - {S_{n - 1}}$ is:
A) ${t_1} + {t_2} + \ldots \ldots + {t_{n - 1}}$
B) ${S_{n - 1}}$
C) \[\mathop \sum \limits_{n = 1}^{n - 2} {t_n}\]
D) ${t_n}$

Answer 1

Hint: First use the formula of the sum of first n terms to find the value of ${S_n}$ and ${S_{n - 1}}$, after that find the difference as required in the problem and then simplify to approach the required result.
Here we will use the formula of the sum of $n$ terms in an arithmetic sequence which is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ where, $a$ represents the first term of the series, $d$ represents the common difference of the series.

Complete step-by-step answer:
It is given that the sequence ${S_n}$ is in arithmetic progression and we have to find ${S_n} - {S_{n - 1}}$ , where ${S_n}$ represents the sum to $n$ terms.
We will apply the formula of the sum of $n$ terms in an arithmetic sequence which is
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
where $a$ represents the first term of the series and $d$ represents the common difference of the series.
So, here is the formula of the sum of $n$ terms of the series of an arithmetic sequence.
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ … (1)
Similarly, we will apply the formula for the sum of $\left( {n - 1} \right)$ terms of the series.
${S_{n - 1}} = \dfrac{{n - 1}}{2}\left[ {2a + \left( {n - 2} \right)d} \right]$ … (2)
Now, we subtract equation (2) from the equation (1) to get the result which is asked in the equation.
$
  {S_n} - {S_{n - 1}} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] - \dfrac{{n - 1}}{2}\left[ {2a + \left( {n - 2} \right)d} \right] \\
   {S_n} - {S_{n - 1}}= \dfrac{1}{2}\left( {2an + {n^2}d - nd - 2an + 2a - {n^2}d + 3nd - 2d} \right)
\\
 {S_n} - {S_{n - 1}} = \dfrac{1}{2}\left( {2a + 2nd - 2d} \right) \\
  {S_n} - {S_{n - 1}} = a + \left( {n - 1} \right)d \\
$
And we know that the formula of ${n^{th}}$ terms of an arithmetic progression is given as:
${T_n} = a + \left( {n - 1} \right)d$ where $a$ represents the first term and $d$ represents the common difference.
And we can clearly observe that the obtained result ${S_n} - {S_{n - 1}}$ is equal ${T_n}$ .
Hence, ${S_n} - {S_{n - 1}} = {T_n}$
Therefore option (D) is correct.

Note: We have to find here the difference that is ${S_n} - {S_{n - 1}}$ by step by step but it can be observed directly in mind because ${S_n}$ represents the sum of all the terms in series and ${S_{n - 1}}$ represents the sum of all the terms in series except the last term. Therefore the result will always be the last term that is ${T_n}$ .