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



3 Comments

Leave a Reply

Contact Us

Call us: 8287971571,0261-4890014

Or, Fill out the form & get a call back.!