Sequence and Series
2.0 Arithmetic Sequence or Arithmetic Progression (A.P.)
2.0 Arithmetic Sequence or Arithmetic Progression (A.P.)
It is a sequence in which the difference between any term and its just preceding term is constant throughout. This constant is called the common difference $(d)$ of the A.P.
If sequence $\{ {t_1},{t_2},{t_3},...\} $ is such that ${t_n} - {t_{n - 1}} = {\text{constant }}\forall n \in N$, then it is an A.P.
How to represent A.P.
First term $=a$, Common difference $=d$
A.P. is written in terms of $a$ and $d$ as $$a,a + d,a + 2d,...,a + (n - 1)d,...$$
- $n$th term of A.P. i.e., $${T_n} = a + (n - 1)d$$ where $d = {T_n} - {T_{n - 1}}$
- $n$th term of A.P. from last i.e., $${T'}_n = l - (n - 1)d$$ where $l$ is the last term.
- Sum of first $n$ terms of A.P. i.e., $${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right] = \frac{n}{2}(a + l)$$ and $${T_n} = {S_n} - {S_{n - 1}}$$
Proof: Sum of first $n$ terms of an A.P. can be written as $${S_n} = {T_1} + {T_2} + {T_3} + ... + {T_n}...(1)$$
It can also be written as $${S_n} = {T_n} + {T_{n - 1}} + {T_{n - 2}} + ... + {T_1}...(2)$$
Adding equations $(1)$ and $(2)$, we get $$2{S_n} = ({T_1} + {T_n}) + ({T_2} + {T_{n - 1}}) + ({T_3} + {T_{n - 2}}) + ... + ({T_n} + {T_1})$$
Let $k = {T_1} + {T_n}$,
Therefore, $$\begin{equation} \begin{aligned} 2{S_n} = nk \\ {S_n} = \frac{n}{2}[{T_1} + {T_n}] \\ {\text{ = }}\frac{n}{2}[a + a + (n - 1)d] \\ {\text{ = }}\frac{n}{2}[2a + (n - 1)d] = \frac{n}{2}[a + l] \\\end{aligned} \end{equation} $$
Note:
- If a sequence is in A.P., and each term of the sequence is added, subtracted, multiplied or divided by any constant term, say $k$, then the resulting series will also be in A.P.
- If ${a_1},{a_2},{a_3},...$ and ${b_1},{b_2},{b_3},...$ are two A.P.s, then ${a_1} \pm {b_1},{a_2} \pm {b_2},{a_3} \pm {b_3},...$ is also in A.P.
- If ${a_1},{a_2},{a_3},...$ and ${b_1},{b_2},{b_3},...$ are two A.P.s, then ${a_1}{b_1},{a_2}{b_2},{a_3}{b_3},...$ and ${a_1}/{b_1},{a_2}/{b_2},{a_3}/{b_3},...$ are not in A.P.
- If ${a_1},{a_2},{a_3},...$ is in A.P., then $$\begin{equation} \begin{aligned} {a_1} + {a_n} = {a_2} + {a_{n - 1}} = {a_3} + {a_{n - 2}} = ... \\ {\text{and}} \\ {a_r} = \frac{{{a_{r - k}} + {a_{r + k}}}}{2}\forall k,0 \leqslant k \leqslant n - r \\\end{aligned} \end{equation} $$
- $2r+1$ $(r \in N)$ terms in A.P. can be taken as $$a - rd,a - (r - 1)d,...,a - d,a,a + d,...,a + (r - 1)d,a + rd$$ For example: Three terms in A.P. can be taken as $$a - d,a,a + d$$ Five terms in A.P. can be taken as $$a - 2d,a - d,a,a + d,a + 2d$$
- $2r$ $(r \in N)$ terms in A.P. can be taken as $$a - (2r - 1)d,a - (2r - 3)d,...,a - 3d,a - d,a + d,a + 3d,...,a + (2r - 3)d,a + (2r - 1)d$$ For example: Four terms in A.P. can be taken as $$a - 3d,a - d,a + d,a + 3d$$ Six terms in A.P. can be taken as $$a - 5d,a - 3d,a - d,a + d,a + 3d,a + 5d$$
- The common difference of A.P. can be zero, positive or negative.
Question 3. Which term is zero in the given sequence $$84,80,76,...$$
Solution: $a=84$, $d=80-84=-4$
Let us assume that the $n$th term is 0. Therefore, $$\begin{equation} \begin{aligned} {T_n} = a + (n - 1)d \\ 0 = 84 + (n - 1) \times - 4 \\ 21 = n - 1 \\ n = 22 \\\end{aligned} \end{equation} $$
$22$nd term is equal to zero in the given sequence.
Question 4. Which term of the sequence $$24,23\frac{1}{4},23\frac{1}{2},21\frac{3}{4},...$$ is the first negative term?
Solution: $a=24$, $d = \frac{{93}}{4} - 24 = - \frac{3}{4}$
Let us assume that the $n$th term is first negative term. Therefore, $$\begin{equation} \begin{aligned} {T_n} < 0 \\ a + (n - 1)d < 0 \\ 24 + (n - 1) \times - \frac{3}{4} < 0 \\ n - 1 > 32 \\ n > 33 \\\end{aligned} \end{equation} $$
$34$th term is the first negative term in the given sequence.
Question 5. $2,5,8,11,...,{T_{50}}$ and $3,5,7,9,11,...,{T_{60}}$ are two A.P.s. Find the number of common terms in both the A.P.s.
Solution: $$\begin{equation} \begin{aligned} {T_{50}} = a + (50 - 1)d = 2 + 49 \times 3 = 149 \\ {T_{60}} = {a'} + (60 - 1){d'} = 3 + 59 \times 2 = 121 \\\end{aligned} \end{equation} $$
So, the last common term of both the A.P.s should be smaller than or equal to $121$. A.P. formed by the common terms of both the A.P.s is $$5,11,17,...$$
From the above A.P., we get $a''=5$, $d''=11-5=6$
$$\begin{equation} \begin{aligned} 121 \geqslant a'' + (n - 1)d'' \\ 121 \geqslant 5 + (n - 1)6 \\ 19 \geqslant n - 1 \\ 20 \geqslant n \\\end{aligned} \end{equation} $$
Therefore, number of common terms is $20$.
Question 6. ${m_{th}}$ term of an A.P. is $\frac{1}{n}$ and ${n_{th}}$ term of an A.P. is $\frac{1}{m}$. Find the $mn$th term i.e., ${T_{mn}}$.
Solution: $$\begin{equation} \begin{aligned} {T_m} = \frac{1}{n} = a + (m - 1)d...(1) \\ {T_n} = \frac{1}{m} = a + (n - 1)d...(2) \\\end{aligned} \end{equation} $$
Subtracting $(2)$ from $(1)$, we get $$\begin{equation} \begin{aligned} a + (m - 1)d - a - (n - 1)d = \frac{1}{n} - \frac{1}{m} \\ d(m - n) = \frac{{m - n}}{{mn}} \\ d = \frac{1}{{mn}} \\ \therefore a = \frac{1}{n} - (m - 1)\left( {\frac{1}{{mn}}} \right) \\ {\text{ = }}\frac{1}{{mn}} \\\end{aligned} \end{equation} $$
$${T_{mn}} = a + (mn - 1)d = \frac{1}{{mn}} + (mn - 1)\left( {\frac{1}{{mn}}} \right) = 1$$
Question 7. ${a_1},{a_2},{a_3},...,{a_{18}}$ is in A.P. and ${a_1} + {a_5} + {a_{10}} + {a_{14}} + {a_9} + {a_{18}} = 220$. Find the value of ${a_2} + {a_{17}}$.
Solution: We know that $${a_1} + {a_{18}} = {a_2} + {a_{17}} = {a_3} + {a_{16}} = {a_4} + {a_{15}} = {a_5} + {a_{14}} = {a_6} + {a_{13}} = {a_7} + {a_{12}} = {a_8} + {a_{11}} = {a_9} + {a_{10}}$$
It is given that $${a_1} + {a_5} + {a_{10}} + {a_{14}} + {a_9} + {a_{18}} = 220$$ or, $$\begin{equation} \begin{aligned} ({a_1} + {a_{18}}) + ({a_5} + {a_{14}}) + ({a_{10}} + {a_9}) = 220 \\ ({a_2} + {a_{17}}) + ({a_2} + {a_{17}}) + ({a_2} + {a_{17}}) = 220 \\ 3{a_2} + 3{a_{17}} = 220 \\ {a_2} + {a_{17}} = 73.3 \\\end{aligned} \end{equation} $$
Question 8. Prove that the sequence is in A.P. if sum of its $n$ terms is of the form $A{n^2} + Bn$.
Solution: As we know that $$\begin{equation} \begin{aligned} {T_n} = {S_n} - {S_{n - 1}} = [A{n^2} + Bn] - [A{(n - 1)^2} + B(n - 1)] \\ {T_n}{\text{ = }}B + A(2n - 1) \\\end{aligned} \end{equation} $$
So, $$\begin{equation} \begin{aligned} {T_1} = B + A(2 - 1) = B + A \\ {T_2} = B + A(2 \times 2 - 1) = B + 3A \\ {T_3} = B + A(2 \times 3 - 1) = B + 5A \\\end{aligned} \end{equation} $$
If it is an A.P., $$\begin{equation} \begin{aligned} {T_2} = \frac{{{T_1} + {T_3}}}{2} \\ LHS \Rightarrow B + 3A \\ RHS \Rightarrow \frac{{2B + 6A}}{2} = B + 3A \\ \therefore LHS = RHS \\ {\text{Hence proved}}{\text{.}} \\\end{aligned} \end{equation} $$
