Arithmetic Progressions

1.0Sequence

In our daily life, we come across the arrangement of numbers or objects in an order such as arrangement of students in a row as per their roll numbers, arrangement of books in the library, etc. An arrangement of numbers depends on the given rule

Thus, a sequence is an ordered arrangement of numbers according to a given rule. Terms of a sequence : The individual numbers that form a sequence are the terms of a sequence. For example : 2, 4, 6, 8, 10,.... forming a sequence are called the first, second, third, fourth, fifth,.... terms of the sequence. The terms of a sequence in successive order is denoted by ' ${\mathrm{T}}_{\mathrm{n}}$ ' or ' ${\mathrm{a}}_{\mathrm{n}}$ '. The ${\mathrm{n}}^{\text{th }}$ term ' ${\mathrm{T}}_{\mathrm{n}}$ ' is called the general term of the sequence. Series The sum of terms of a sequence is called the series of the corresponding sequence. ${\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3+\dots$. is an infinite series, where as $T_1+T_2+T_3+\dots +T_{n-1}+T_n$ is a finite series of $n$ terms. Usually, the series of finite number of $n$ terms is denoted by ${\mathrm{S}}_{\mathrm{n}}$. ${\mathrm{S}}_{\mathrm{n}}={\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3+\dots +{\mathrm{T}}_{\mathrm{n}-2}+{\mathrm{T}}_{\mathrm{n}-1}+{\mathrm{T}}_{\mathrm{n}}$

Numerical Ability 1 Write the first five terms of the sequence, whose $n^{\text{th }}$ term is $a_n=\left\{1+(-1{)}^n\right\}n$. Solution: $a_n=\left\{1+(-1{)}^n\right\}n$ Substituting $\mathrm{n}=1,2,3,4$ and 5 , we get $a_1=\left\{1+(-1{)}^1\right\}1=0;a_2=\left\{1+(-1{)}^2\right\}2=4$; ${\mathrm{a}}_3=\left\{1+(-1{)}^3\right\}3=0;{\mathrm{a}}_4=\left\{1+(-1{)}^4\right\}4=8$; ${\mathrm{a}}_5=\left\{1+(-1{)}^5\right\}5=0$ Thus, the required terms are : $0,4,0,8$ and 0 .

2.0Progression

It is not always possible to write each and every sequence of some rule.

3.0Arithmetic progressions

• We know about some sequences which are arithmetic progressions as. (i) Sequence of even numbers: $2,4,6,8,10$, (ii) Sequence of odd numbers: $1,3,5,7,9$, ... (iii) Sequence of multiples of 5: $5,10,15,20,25$, (iv) Sequence of number which are divisible by 7 and give remainders 1 in each case: $8,15,22,29,36$,
  

An arithmetic progression is list of numbers (terms) in which the first term is given and each term, other than the first term is obtained by adding a fixed number 'd' to the preceding term.

The fixed number ' d ' is known as the common difference of the arithmetic progression. Its value can be positive, negative or zero.

The first term is denoted by 'a' or 'a1' and the last term by ' $\ell$ '. e.g., Consider a sequence $6,10,14,18,22$, .....

Here, $a_1=6,a_2=10,a_3=14,a_4=18,a_5=22$ $a_2-a_1=10-6=4$ ${\mathrm{a}}_3-{\mathrm{a}}_2=14-10=4$ ${\mathrm{a}}_4-{\mathrm{a}}_3=18-14=4$

A sequence of non-zero numbers ${\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,..,{\mathrm{a}}_{\mathrm{n}}$ is said to be a geometric sequence or GP, if $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\cdots$ i.e. if $\frac{a_{n+1}}{a_n}=a$ constant for all n. e.g., 3 , $9,27,81,\dots$ SPOT LIGHT

Therefore, the sequence is an arithmetic progression in which the first term $\mathrm{a}=6$ and the common difference $\mathrm{d}=4$.

Symbolic form : Let us denote the first term of an AP by $a_1$, second term by $a_2,\dots ..n^{\text{th }}$ term by $a_n$ and the common difference by $d$. Then the AP becomes $a_1,a_2,a_3,\dots ,a_n$.

So, ${\mathrm{a}}_2-{\mathrm{a}}_1={\mathrm{a}}_3-{\mathrm{a}}_2=\dots ={\mathrm{a}}_{\mathrm{n}}-{\mathrm{a}}_{\mathrm{n}-1}=\mathrm{d}$. General form : In general form, an arithmetic progression with first term 'a' and common difference ' $d$ ' can be represented as follows: $a,a+d,a+2d,a+3d,a+4d,\dots$.

Finite AP : An AP in which there are only a finite number of terms is called a finite AP. It may be noted that each such AP has a last term.

e.g., (a) The heights (in cm) of some students of a school standing in a queue in the morning assembly are $147,148,149,\dots ,157$. (b) The minimum temperatures (in degree Celsius) recorded for a week in the month of January in a city arranged in ascending order are $-3.1,-3.0,-2.9,-2.8,-2.7,-2.6,-2.5$

Infinite AP : An AP in which the number of terms is not finite is called infinite AP. That means infinite AP does not have a last term.

The sum of cubes of first n natural numbers i.e., $1^3+2^3+3^3+\dots +n^3$ is usually written as $\Sigma n^3$. $\sum n^3={\left(\frac{n(n+1)}{2}\right)}^2={\left(\sum n\right)}^2$

e.g., (a) $1,2,3,4,\dots$ (b) $100,70,40,10\dots$

Least information required : To know about an AP, the minimum information we need to know is the first term a and the common difference d. For instance if the first term a is 6 and the common difference $d$ is 3 , then AP is $6,9,12,15,\dots$. Similarly, when $a=-7,d=-2$, the AP is $-7,-9,-11,-13,\dots$ $\mathrm{a}=1.0,\mathrm{d}=0.1$, the AP is $1.0,1.1,1.2,1.3,\dots$ So, if we know what a and d are, we can list the AP.

• If a constant is added or subtracted from each term of an AP then the resulting sequence is also an AP with the same common difference.

In which of the following situations, the list of numbers obtained will be in the form of an arithmetic progression? (i) Number of students left in the school auditorium from the total strength of $1000$ students when they leave the auditorium in batches of 25 . (ii) The amount of air present in the cylinder when a vacuum pump removes each time $1/3$ of the air left in the cylinder. (iii) Cash price of a particular brand of washing machine in the market is $₹12000$. Sawitri buys one washing machine on monthly instalments of $₹1000$ plus an interest at the rate of $2\%$ per month on the balance amount. She makes the first instalment after one month. No amount is to be paid at the time of purchase. Explanation: (i) ${\mathrm{t}}_1=1000,{\mathrm{t}}_2=1000-25=975,{\mathrm{t}}_3=975-25=950,{\mathrm{t}}_4=950-25=925$ and so on. Thus, the list of numbers is as below : 1000, 975, 950, 925, ...... Here, ${\mathrm{t}}_2-{\mathrm{t}}_1={\mathrm{t}}_3-{\mathrm{t}}_2={\mathrm{t}}_4-{\mathrm{t}}_3=\dots ...=-25$. Therefore, the above list of numbers forms an AP (ii) Let us suppose that air present in the cylinder in the beginning is x units. Every time the vacuum pump removes the $\frac{1}{3}$ of the air present in the cylinder. Here, ${\mathrm{t}}_1=\mathrm{x}$ units, ${\mathrm{t}}_2=\left(\mathrm{x}-\frac{1}{3}\mathrm{x}\right)$ units $=\frac{2}{3}\mathrm{x}$ units, ${\mathrm{t}}_3=\left\{\frac{2}{3}\mathrm{x}-\frac{1}{3}\left(\frac{2}{3}\mathrm{x}\right)\right\}$ units $=\left\{\frac{2}{3}\mathrm{x}-\frac{2}{9}\mathrm{x}\right\}$ units $=\frac{6x-2x}{9}$ units $=\frac{4}{9}\mathrm{x}$ units, ${\mathrm{t}}_4=\left\{\frac{4}{9}\mathrm{x}-\frac{1}{3}\left(\frac{4}{9}\mathrm{x}\right)\right\}$ units $=\left\{\frac{4}{9}\mathrm{x}-\frac{4}{27}\mathrm{x}\right\}$ units $=\frac{8}{27}\mathrm{x}$ units and so on.

In a finite AP the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of first and last term.

Thus the list of numbers is as below : $\mathrm{x},\frac{2}{3}\mathrm{x},\frac{4}{9}\mathrm{x},\frac{8}{27}\mathrm{x},\dots \dots \dots$ Here, ${\mathrm{t}}_2-{\mathrm{t}}_1=\frac{2}{3}\mathrm{x}-\mathrm{x}=-\frac{1}{3}\mathrm{x}$ ${\mathrm{t}}_3-{\mathrm{t}}_2=\frac{4}{9}\mathrm{x}-\frac{2}{3}\mathrm{x}=\frac{4\mathrm{x}-6\mathrm{x}}{9}=-\frac{2}{9}\mathrm{x}$ ${\mathrm{t}}_4-{\mathrm{t}}_3=\frac{8}{27}\mathrm{x}-\frac{4}{9}\mathrm{x}=\frac{8\mathrm{x}-12\mathrm{x}}{27}=-\frac{4}{27}\mathrm{x}$ and so on. i.e., ${\mathrm{t}}_2-{\mathrm{t}}_1\ne {\mathrm{t}}_3-{\mathrm{t}}_2\ne {\mathrm{t}}_4-{\mathrm{t}}_3$. Obviously, the common differences of two consecutive terms are not same throughout. Hence, the list of numbers does not form an AP. (iii) ${\mathrm{t}}_1=₹1000+₹\left(12000\times \frac{2}{100}\right)=₹1000+₹240=₹1240$, ${\mathrm{t}}_2=₹1000+₹\left(11000\times \frac{2}{100}\right)=₹1000+₹220=₹1220$, ${\mathrm{t}}_3=₹1000+₹\left(10000\times \frac{2}{100}\right)=₹1000+₹200=₹1200$, ${\mathrm{t}}_4=₹1000+₹\left(9000\times \frac{2}{100}\right)=₹1000+₹180=₹1180$ and so on. Thus, the list of the amounts of instalments is (in rupees) 1240, 1220, 1200, 1180, .... ${\mathrm{t}}_2-{\mathrm{t}}_1=1220-1240=-20$ ${\mathrm{t}}_3-{\mathrm{t}}_2=1200-1220=-20$ ${\mathrm{t}}_4-{\mathrm{t}}_3=1180-1200=-20$ and so on. Hence, the list form an AP.

Numerical Ability 2 Write first four terms of an AP, when the first term a and the common difference $d$ are as follows : $a=4,d=5$. Solution: First term, $\mathrm{a}=4$ Second term $=4+d=4+5=9$ Third term $=9+d=9+5=14$ Fourth term $=14+\mathrm{d}=14+5=19$ Hence, first four terms of the given AP are 4, 9, 14, 19. General term of an arithmetic progression The formula for writing general term or the ${\mathrm{n}}^{\text{th }}$ term of

r^{\text {th }} term of a finite arithmetic progression from the end Let there be an arithmetic progression with first term a and common difference d. If there are $n$ terms in the arithmetic progression, then $r^{\text{th }}$ term from the end $=a+(n-r)d$

• If $\ell$ is the last term of an arithmetic progression then $r^{\text{th }}$ term from the end is the $r^{\text{th }}$ term of an arithmetic progression whose first term is $\ell$ and common difference is -d . ${\mathrm{r}}^{\text{th }}$ term from the end $=\ell +(\mathrm{r}-1)(-\mathrm{d})$
• Whenever we find number of terms(n) for any arithmetic progression, then n should be natural number only.

Numerical Ability 3 Find the $20^{\text{th }}$ term of the sequence : 7, 3, $-1,-5,\dots$. Solution ${\mathrm{t}}_2-{\mathrm{t}}_1=3-7=-4$ ${\mathrm{t}}_3-{\mathrm{t}}_2=-1-3=-4$ ${\mathrm{t}}_4-{\mathrm{t}}_3=-5-(-1)=-4$ The given sequence is an AP in which the first term $\mathrm{a}=7$ and the common difference $\mathrm{d}=-4$. ${\mathrm{t}}_{20}=\mathrm{a}+(20-1)\mathrm{d}\left\{\because {\mathrm{t}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}+1)\mathrm{d}\right\}$ $=7+(20-1)\times (-4)$ $=7-4\times 19=7-76=-69$ Hence, ${\mathrm{t}}_{20}=-69$.

Numerical Ability 4 Find the $6^{\text{th }}$ term from the end of the AP 17, 14, 11, ...., $-40$. Solution: The given AP : 17, 14, 11, ..., -40 Here, $a=17,d=14-17=-3,\ell =-40$ Let there be $n$ terms in the given AP. Then, ${\mathrm{n}}^{\text{th }}$ term $=-40$ $\Rightarrow a+(n-1)d=-40\left(\because a_n=a+(n-1)d\right)$ $\Rightarrow 17+(\mathrm{n}-1)(-3)=-40$ $\Rightarrow (\mathrm{n}-1)(-3)=-40-17$ $\Rightarrow (\mathrm{n}-1)(-3)=-57$ $\Rightarrow \mathrm{n}-1=\frac{-57}{-3}$ $\Rightarrow \mathrm{n}-1=19$ $\Rightarrow \mathrm{n}=19+1$ $\Rightarrow \mathrm{n}=20$ Hence, there are 20 terms in the given AP. Now, 6th term from the end $=a+(20-6)d(\because r^{\text{th }}$ term from the end $=a+(n-r)d)$ $=\mathrm{a}+14\mathrm{d}$ $=17+14(-3)$ $=17-42=-25$ Hence, the $6^{\text{th }}$ term from the end of the given AP is -25 . Alternate solution $6^{\text{th }}$ term from end $=\ell +(\mathrm{n}-1)(-\mathrm{d})$ $=-40+(6-1)(-(-3))$ $=-40+5(3)$ $=-40+15$ $=-25$

A sequence is an AP if the sum of its first $n$ terms is of the form ${\mathrm{An}}^2+\mathrm{Bn}$, where A, B are constants independent of $n$. In such a case, the common difference is 2A.

4.0Selection of terms in an AP

Sometimes we require certain number of terms in AP whose sum is given. The following ways of selecting terms are generally very convenient.

It should be noted that in case of an odd number of terms, the middle term is a and the common difference is $d$ while in case of an even number of terms the middle terms are $a-d,a+d$ and the common difference is 2 d .

• Selection of terms is useful when sum of certain terms are given. If the sum of terms is not given, then select terms as $\mathrm{a},\mathrm{a}+\mathrm{d},\mathrm{a}+2\mathrm{d}\dots \dots .$.
• If three numbers $a,b,c$ in order are in AP. Then $\mathrm{b}-\mathrm{a}=$ Common difference $=\mathrm{c}-\mathrm{b}$ $\Rightarrow \mathrm{b}-\mathrm{a}=\mathrm{c}-\mathrm{b}$ $\Rightarrow 2b=a+c$ Thus, $a,b,c$ are in AP if and only if $2b=a+c$.
• If $a,b,c$ are in AP, then $b$ is known as the arithmetic mean (AM) between $a$ and $c$.
• If $a,x,b$ are in AP then, $2x=a+b$ $\Rightarrow \mathrm{x}=\frac{\mathrm{a}+\mathrm{b}}{2}$. Thus, AM between a and b is $\frac{\mathrm{a}+\mathrm{b}}{2}$.

Numerical Ability 5 The sum of three numbers in AP is $-3$, and their product is 8 . Find the numbers. Solution: Let the three numbers in AP be $(a-d),a,(a+d)$. Then, $\Rightarrow$ Sum $=-3$ $\Rightarrow (\mathrm{a}-\mathrm{d})+\mathrm{a}+(\mathrm{a}+\mathrm{d})=-3$ $\Rightarrow 3a=-3$ $\Rightarrow \mathrm{a}=-1$ Now, product $=8$ $\Rightarrow (\mathrm{a}-\mathrm{d})(\mathrm{a})(\mathrm{a}+\mathrm{d})=8$ $\Rightarrow a\left(a^2-d^2\right)=8$ $\Rightarrow (-1)\left(1-{\mathrm{d}}^2\right)=8(\because a=-1)$ $\Rightarrow {\mathrm{d}}^2=9$ $\Rightarrow \mathrm{d}=\pm 3$ If $d=3$, the numbers are $-4,-1,2$. If $d=-3$, the numbers are $2,-1,-4$. Thus, the numbers are $-4,-1,2$ or $2,-1,-4$.

Numerical Ability 6 If $2x,x+10,3x+2$ are in AP, find the value of $x$. Solution: As, $2x,x+10,3x+2$ are in AP, $\therefore 2(x+10)=2x+(3x+2)$ $\Rightarrow 2x+20=5x+2$ $\Rightarrow 3\mathrm{x}=18$ $\Rightarrow x=6$

5.0Sum of first \mathbf{n} terms of an arithmetic progression

The sum ' ${\mathrm{S}}_{\mathrm{n}}$ ' of n terms of an arithmetic progression with first term 'a' and common difference ' d ' is ${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2\mathrm{a}+(\mathrm{n}-1)\mathrm{d}]$ or ${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[\mathrm{a}+\ell ];$ Where $\ell =$ last term.

• In the formula $S_n=\frac{n}{2}[2a+(n-1)d]$, there are four quantities viz. $S_n,a,n$ and d. If any three of these are known, the fourth can be determined. Sometimes, two of these quantities are given. In such a case, remaining two quantities are provided by some other relation.
• If the sum $S_n$ of $n$ terms of a sequence is given, then $n^{\text{th }}$ term $a_n$ of the sequence can be determined by using the following formula $a_n=S_n-S_{n-1}$ i.e., the $n^{\text{th }}$ term of an AP is the difference of the sum of first $n$ terms and the sum of first ( $n-1$ ) terms of it.

Numerical Ability 7 Find the sum of the first 20 terms of the AP : 5, 8, 11, 14, ..... . Solution First term of the $\mathrm{AP}=5$, i.e., $\mathrm{a}=5$ Common difference $=3$, i.e., $\mathrm{d}=3$ and $\mathrm{n}=20$ $S_{20}=\frac{20}{2}(2a+19d)=10\times (10+19\times 3)=10\times 67=670$

Numerical Ability 8 Find the sum : $34+32+30+\dots .+10$ Solution: As the given AP is $34+32+30+\dots .+10$ Here, $\mathrm{a}=34$ $$\begin{gathered} \mathrm{d}=32-34=-2 \\ \ell =10 \end{gathered}$$ Let the number of terms of the AP be $n$. We know that $$\begin{gathered} {\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d} \\ \Rightarrow 10=34+(\mathrm{n}-1)(-2) \\ \Rightarrow (\mathrm{n}-1)(-2)=-24 \\ \Rightarrow \mathrm{n}-1=\frac{-24}{-2}=12 \\ \Rightarrow \mathrm{n}=13 \\ \text{ Again, we know that, } \\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\ell ) \\ \Rightarrow {\mathrm{S}}_{13}=\frac{13}{2}(34+10) \\ \Rightarrow {\mathrm{S}}_{13}=286 \end{gathered}$$ Hence, the required sum is 286.

Numerical Ability 9 Find the number of terms of the AP 54, 51, 48,...so that their sum is 513. Solution: The given AP is $54,51,48,\dots$ Here, $a=54,d=51-54=-3$ Let the sum of $n$ terms of this AP be $S_n$. We know that $S_n=\frac{n}{2}[2a+(n-1)d]$ $\Rightarrow 513=\frac{\mathrm{n}}{2}[2(54)+(\mathrm{n}-1)(-3)]$ $\Rightarrow 513=\frac{\mathrm{n}}{2}[108-3\mathrm{n}+3]$ $\Rightarrow 513=\frac{\mathrm{n}}{2}[111-3\mathrm{n}]$ $\Rightarrow 1026=\mathrm{n}[111-3\mathrm{n}]$ $\Rightarrow 1026=111\mathrm{n}-3{\mathrm{n}}^2$ $\Rightarrow 3n^2-111n+1026=0$ $\Rightarrow {\mathrm{n}}^2-37\mathrm{n}+342=0$ (Dividing throughout by 3 ) $\Rightarrow n^2-18n-19n+342=0$ $\Rightarrow \mathrm{n}(\mathrm{n}-18)-19(\mathrm{n}-18)=0$ $\Rightarrow (\mathrm{n}-18)(\mathrm{n}-19)=0$ $\Rightarrow \mathrm{n}-18=0$ or $\mathrm{n}-19=0$ $\Rightarrow \mathrm{n}=18,19$ Hence, the sum of 18 terms or 19 terms of the given AP is 513. So, $\mathrm{n}=18$ or 19 Note: Here 19th term $={\mathrm{a}}_{19}$ ${\mathrm{a}}_{19}=\mathrm{a}+(19-1)\mathrm{d}$ $\left(\because a_n=a+(n-1)d\right)$ ${\mathrm{a}}_{19}=\mathrm{a}+18\mathrm{d}$ ${\mathrm{a}}_{19}=54+18(-3)$ ${\mathrm{a}}_{19}=54-54=0$

Numerical Ability 10 Rakesh has to buy a TV. He can buy TV either making cash down payment of ₹ $15000$ at once or by making 14 monthly instalments as below : ₹ 1500 ( $1^{\text{st }}$ month), ₹ 1450 ( $2^{\text{nd }}$ month), ₹ 1400 ( 3 rd month), ₹ 1350 ( 4 th month), ..... Each instalment except the first is ₹ $50$ less than the previous one. Find: (i) Amount of the instalment paid in the $9^{\text{th }}$ month. (ii) Total amount paid in 14 instalments. (iii) How much extra he has to pay in addition to the amount of cash down payment? Solution: ${\mathrm{t}}_1=1500,{\mathrm{t}}_2=1450,{\mathrm{t}}_3=1400,{\mathrm{t}}_4=1350,\dots ..$. Now, $t_2-t_1=t_3-t_2=\dots =-50$. So, the fourteen monthly instalments form an AP. Here, $\mathrm{a}=1500$ and $\mathrm{d}=-50$ (i) ${\mathrm{t}}_9=\mathrm{a}+8\mathrm{d}=1500-8\times 50=1500-400$ Thus, $9^{\text{th }}$ instalment $=₹1100$ (ii) ${\mathrm{S}}_{14}=\frac{14}{2}(2\mathrm{a}+13\mathrm{d})=7\times (2\times 1500+13\times -50)$ $=7\times (3000-650)=7\times 2350=16,450$ Here, total amount paid = ₹ 16,450 (iii) Extra paid amount (i.e., in addition to cash down payment) $=₹16,450-₹15,000=₹1450$