Question

Which of the following formula is true for ${T_n}$
A.${T_n} = {S_n} + {S_{n - 1}}$
B.${T_n} = {S_n} - {S_{n + 1}}$
C.${T_n} = {S_n} - {S_{n - 1}}$
D.${T_n} = {S_{n - 1}} - {S_n}$

Answer 1

Here, we will find the relationship between the arithmetic series and the arithmetic progression. Arithmetic Progression is a sequence of numbers such that the common difference between two consecutive terms is a constant. The arithmetic series is the sum of the terms of the arithmetic progression.

Formula Used:
We will use the following formulas:
1.Arithmetic Series is given by the formula ${S_n} = \dfrac{n}{2}\left( {a + \left( {n - 1} \right)d} \right)$
2.Arithmetic Progression is given by the formula ${T_n} = a + \left( {n - 1} \right)d$
where ${S_n}$ is the sum of n terms of an arithmetic progression, ${T_n}$is the $n$ th term of the series, $a$ is the first term of the series, $d$ is the common difference which is a constant.

Complete step-by-step answer:
Let us consider a series of terms which are in A.P.
Let ${S_n},{S_{n - 1}}$ be the sum of the arithmetic progression of $n,n - 1$ terms respectively.
Let ${T_n}$ be the arithmetic progression of n terms.
Arithmetic Progression is given by the formula ${T_n} = a + \left( {n - 1} \right)d$…………………………………………..$\left( 1 \right)$
Arithmetic Series is given by the formula ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
The sum of $n$ terms which are in A.P. is given by ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$……………………………………….$\left( 2 \right)$
The sum of $n - 1$ terms which are in A.P. is given by ${S_{n - 1}} = \dfrac{{n - 1}}{2}\left( {2a + \left( {\left( {n - 1} \right) - 1} \right)d} \right)$……………………$\left( 3 \right)$
By subtracting equations $\left( 2 \right)$and $\left( 3 \right)$, we get
${S_n} - {S_{n - 1}} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) - \dfrac{{n - 1}}{2}\left( {2a + \left( {\left( {n - 1} \right) - 1} \right)d} \right)$
By rewriting the equation, we get
$\Rightarrow {S_n} - {S_{n - 1}} = \dfrac{1}{2}\left( {2an + n\left( {n - 1} \right)d} \right) - \dfrac{1}{2}\left( {2a\left( {n - 1} \right) + \left( {\left( {n - 1} \right)\left( {n - 2} \right)} \right)d} \right)$
By grouping the terms, we get
$\Rightarrow {S_n} - {S_{n - 1}} = \dfrac{1}{2}\left( {2a\left( {n - n + 1} \right) + \left( {n - 1} \right)\left( {n - n + 2} \right)d} \right)$
By rewriting the equation, we get
$\Rightarrow {S_n} - {S_{n - 1}} = \dfrac{1}{2}\left( {2a + 2\left( {n - 1} \right)d} \right)$
Taking 2 common from the expression on the RHS, we get
$\Rightarrow {S_n} - {S_{n - 1}} = \dfrac{2}{2}\left( {a + \left( {n - 1} \right)d} \right)$
Dividing the terms, we get
$\Rightarrow {S_n} - {S_{n - 1}} = \left( {a + \left( {n - 1} \right)d} \right)$
Substituting ${T_n} = a + \left( {n - 1} \right)d$ in the above equation, we get
$\Rightarrow {S_n} - {S_{n - 1}} = {T_n}$
Therefore, ${T_n} = {S_n} - {S_{n - 1}}$.
Thus Option(C) is the correct answer.

Note: If the same number is added or subtracted from each term of an A.P., then the resulting terms in the sequence are also in A.P. but with the same common difference. Arithmetic sequences may be finite or infinite. In order to solve this question, we need to keep in mind the formula for both arithmetic series and arithmetic progression. We might make a mistake by using the formula of geometric progression instead of an arithmetic progression. Geometric progression is a sequence where the consecutive terms differ by a common ratio.