# Arithmetic Progression - Class 10 : Notes

Notes for arithmetic progression chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

(1) A sequence is an arrangement of numbers or objects in a definite order.
For Example: 1, 8, 27, 64, 125,……
Above arrangement numbers are arranged in a definite order according to some rule.

(2) A sequence ${a_1},{a_2},{a_3},....,{a_n},..$ is called an arithmetic progression, if there exists a constant d such that, ${a_2} - {a_1} = d,{a_3} - {a_2} = d,{a_4} - {a_3} = d,...,{a_{n + 1}}-{a_n} = d$ and so on. The constant d is called the common difference.
For Example: 2, 4, 6, 8,…. is a arithmetic progression because number are even natural numbers where ${a_1} = 2,{a_2} = 4,{a_3} = 6,{a_4} = 8$
$4- 2 = 2, 6 - 4 = 2, {8} - {6} = d,...,{a_{n + 1}} - {a_n} = 2$

(3) If ‘a’ is the first term and 'd' the common difference of an AP,  then the A.P.  is $a,a + d,a + 2d,a + 3d,a + 4d....$
For Example: If AP is 2, 4, 6, 8,…. Then first term $a=2$ and $d=2$
So, $2,2 + 4,2 + 2(2),2 + 3(2),a + 4(2)....$.

(4) A sequence ${a_1},{a_2},{a_3},....,{a_n},..$ is an AP, if ${a_{n + 1}} - {a_n}$ is independent of n.
For Example: If sequence is 2, 4, 6, 8, …… ${a_n}$,….. so if we take ${a_n} = 16$ so ${a_{n + 1}}=18$ So ${a_{n + 1}} - {a_n} = 18 - 16 = 2$ which is independent of n.

(5) A sequence ${a_1},{a_2},{a_3},....,{a_n},..$ is an AP, if and only if its ${n^{th}}$ term ${a_n}$ is a linear expression in n and In such a case the coefficient of n is the common difference.
For Example: A sequence 1, 4, 9, 16, 25,…. Is an AP. Suppose ${n^{th}}$ term ${a_n}= 81$ which is a linear expression in n. which is ${n^2}$.

(6) The ${n^{th}}$ term ${a_n}$, of an AP with first term ‘a’ and common difference ‘d’ is given by ${a_n} = a + (n - 1)d$
For Example: If want to find ${n^{th}}$  term ${a_n}$ in example given in 4th .
$a=2$, $d=2$ then we can find 10th term by putting n=10 in above equation. So 10th term of  sequence is ${a_{10}} = 2 + (10 - 1)2 = 20$

(7) Let there be an A.P with first term ‘a’ and common difference d. if there are m terms in the AP, then
${n^{th}}$ term from the end = ${(m - n + 1)^{th}}$
term from the beginning =$a + (m - n)d$
Also, ${n^{th}}$ term from the end = Last term + $(n - 1)( - d)$

= $l - (n - 1)d$, where $l$ denotes the last term.
For Example: Determine the 10th term from the end of the A.P 4, 9, 14, …, 254.
$l = 254$, $d=5$
${n^{th}}$ term from the end =$l - (10 - 1)d$ = $l - 9d$= $254 - 9 \times 5$=209

(8) Various terms is an AP can be chosen in the following manner.

 Number of terms Terms Common difference 3 $a - d,a,a + d$ d 4 $a - 3d,a - d,a + d,a + 3d$ 2d 5 $a - 2d,a - d,a,a + d,a + 2d$ d 6 $a - 5d,a - 3d,a-d,a + d,a + 3d,a + 5d$ 2d

(9) The sum to n terms of an A.P with first term ‘a’ and common difference ‘d’ is given by ${S_n} = {n \over 2}\left\{ {2a + (n - 1)d} \right\}$ Also, ${S_n} = {n \over 2}\left\{ {a + l} \right\}$, where $l$= last term = $a + (n - 1)d$
For Example: (i) 50, 46, 42, … find the sum of first 10th term
Solution:
Given, $50,46,42,.....$
Here ,  first term $a = 50$,
Difference $d = 46 - 50 =(- 4)$
And no of terms $n = 10$
We know ${S_n} = {n \over 2}\left[ {2a + (n - 1)d} \right]$
${S_n} = {{10} \over 2}\left[ {2(50) + (10 - 1)( - 4)} \right]$ $\Rightarrow 5\left[ {100 + (9)( - 4)} \right]$
${S_n} = 5\left[ {100 - 36} \right] \Rightarrow 5 \times 64 \Rightarrow 320$
Hence, Sum of 10 terms is 320.

(ii) First term is 17 and last term is 350 and d=9 so find total sum and find how many terms are there.
Solution:
Given, first term, a=17, last term, ${a_n}$ = 350 = $l$
And difference d = 9
We know, ${a_n} = a + (n - 1)d$
$350 = 17 + (n - 1)9$
$350 = 17 + 9n - 9$
$9n = 350 - 17 + 9 \Rightarrow 342$
$n = 38$
We know, sum of n terms
${S_n} = {n \over 2}(a + l)$
${S_{38}} = {{38} \over 2}\left[ {17 + 350} \right] \Rightarrow 19 \times 367 \Rightarrow 6973$
Hence, number of terms is 38 and sum is 6973.

(10) If the ratio of the sums of n terms of two AP’s is given, then to find the ratio of their ${n^{th}}$ terms, we replace n by (2n-1) in the ratio of the sums of n terms.
For Example: The ratio of the sum of n terms of two AP’s is (7n+1):(4n+27). Find the ration of their ${m^{th}}$ terms.
Solution:
let ${a_1}$, ${a_2}$ be the 1st terms and ${d_1}$, ${d_2}$ the common differences of the two given A.P’s. then the sums of their n terms are given by,
${S_{n1}} = {n \over 2}\left\{ {2{a_1} + (n - 1){d_1}} \right\}$ and ${S_{n2}} = {n \over 2}\left\{ {2{a_2} + (n - 1){d_2}} \right\}$
${{{S_{n1}}} \over {S_{n2}}} = {{{n \over 2}\left\{ {2{a_1} + (n - 1){d_1}} \right\}} \over {{n \over 2}\left\{ {2{a_2} + (n - 1){d_2}} \right\}}}$
${{{S_{n1}}} \over {S_{n2}}} = {{2{a_1} + (n - 1){d_1}} \over {2{a_2} + (n - 1){d_2}}}$
It is given that ${{{S_{n1}}} \over {S_{n2}}} = {{7n + 1} \over {4n + 27}}$
${{7n + 1} \over {4n + 27}} = {{2{a_1} + (n - 1){d_1}} \over {2{a_2} + (n - 1){d_2}}}$...........(i)

To find ratio of the ${m^{th}}$ terms of the two given AP’s, we replace $n$ by $(2m-1)$ in equation (i). Therefore,
${{7(2m - 1) + 1} \over {4(2m - 1) + 27}} = {{2{a_1} + ((2m-1) - 1){d_1}} \over {2{a_2} + ((2m-1) - 1){d_2}}}$
${{14m - 6} \over {8m + 23}} = {{{a_1} + (m - 1){d_1}} \over {{a_2} + (m - 1){d_2}}}$
Hence, the ratio of the ${m^{th}}$ terms of the two AP’s is ${{14m - 6} \over {8m + 23}}$
So as per rule if we replace $n$ by $(2m-1)$ we get ratio $(14m-6):(8m+23)$

(11) A sequence is an AP if and only if the sum of its n terms is of the form $A{n^2} + Bn$, where A,  B are constants.  In such a case the common difference is 2A.
For Example:
For the A.P
${S_n} = {p_n} + q_n^2$
Now ${S_1} = p \times 1 + q{(1)^2}$
${S_1} = p \times q \Rightarrow {T_1} = p + q$ and also ${S_2} = p \times 2 + q{(2)^2}$
${S_2} = 2p + 4q$
We have ${T_1} + {T_2} = 2p + 4q$
Or ${T_2} = 2p + 4q - {T_1}$
${T_2} = 2p + 4q - (p + q) \Rightarrow p + 3q$
Hence common difference = ${T_2} - {T_1}$
$= p + 3q - (p + q) = 2q$

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