Class 10 Chapter 5 (arithmetic progression) class notes

 

 Introduction to AP

Sequences, Series and Progressions

•     A sequence is a finite or infinite list of numbers following a certain pattern.  For example: 1, 2, 3, 4, 5… is the sequence, which is infinite sequence of natural numbers.

•     A series is the sum of the elements in the corresponding sequence. For example: 1 + 2 + 3 + 4 + 5…. is the series of natural numbers. Each number in a sequence or a series is called a term.

•     A progression is a sequence in which the general term can be can be expressed using a mathematical formula. For Example: 2, 4, 6, 8,…. is a arithmetic progression because number are even natural numbers

 

 >   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....  and so on
    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) .....

>  A sequence $a_1,a_2,a_3\mathrm{.......}{a}_n$ is an AP, if $a_1,a_2,a_3\mathrm{.......}{a}_{n+1}-a_{n}isindependentofn$
    For Example: If sequence is 2, 4, 6, 8, …… $a_n$ ,….. so if we take $a_n=16$ so $a_{n+1}=18$ ,     Here $a_{n+1}-a_n= 18-\mathrm{16 }= 2$    which is independent of n.

>   A sequence $a_1,a_2,a_3\mathrm{.......}{a}_n$ is an AP, if and only if its n^th term an 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 n².

>  The n^th term an, 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 nth  term an in example given in 4th .
    a = 2, d = 2 then we can find 10^th term by putting n = 10 in above equation. So 10^th term of     sequence is a₁₀ = 2 + (10 − 1)2 = 20


>  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 10^th term from the end of the A.P 4, 9, 14, …, 254.
    l = 254, d = 5
    nth term from the end =l − (10 − 1)d = l − 9d = 254 − 9 × 5 = 209

>  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

>  The sum to n terms of an A.P with first term ‘a’ and common difference ‘d’ is given by 

    $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$  

    Also, $S_n=\frac{n}{2}\left[a+l\right]$ , where l = last term = a + (n − 1)d
    For Example: (i) 50, 46, 42, … find the sum of first 10 terms
    Solution:
$$\begin{array}{l}Given,50,46,42,\mathrm{.....}Here,firstterm(a)=50,commondifference(d)=46-50=(-4)\\ Andnooftermsn=10\\ Weknow{S}_n=\frac{n}{2}\left[2a+(n-1)d\right]\\ \Rightarrow S_n=\frac{10}{2}\left[2\times 50+(10-1)(-4\right]\\ \Rightarrow 5\left[100+9\times -4\right]=5\left[100-36\right]\\ \Rightarrow 5\times 64=320\end{array}$$     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, an = 350 = l
    And difference d = 9
    We know, $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$
    350 = 17 + (n − 1)9
    350 = 17 + 9n − 9
    9n = 350 − 17 + 9 ⇒ 342
    n = 38
    We know, sum of n terms
    $S_n=\frac{n}{2}\left[a+l\right]$
    $S_{38}=\frac{38}{2}\left[17+350\right]$ ⇒ 19 × 367 ⇒ 6973
    Hence, number of terms is 38 and sum is 6973.
    

>   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 $\frac{7n+1}{4n+27}$ . Find the ration of their m^th terms.
Solution:
let a1, a2 be the 1st terms and d1, d2 the common differences of the two given A.P’s. then the sums of their n terms are given by,
$$\Rightarrow \frac{S_{n_1}}{S_{n_2}}=\frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]}$$ $$\Rightarrow \frac{S_{n_1}}{S_{n_2}}=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}$$
$$\text{It}\text{is}\text{given}\text{that}\frac{S_{n_1}}{S_{n_2}}=\frac{7n+1}{4n+27}$$
$$\Rightarrow \frac{7n+1}{4n+27}=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}\mathrm{.......}(i)$$

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

>  A sequence is an AP if and only if the sum of its n terms is of the form An²+Bn, where A,  B are     constants.  In such a case the common difference is 2A.                            
    For Example:
    For the A.P
    S_n = pn + qn²
    Now S₁ = p × 1 + q(1)²
    S₁ = p × q ⇒T₁ = p + q and also S₂ = p × 2 + q(2)²
    S₂ = 2p + 4q
    We have T₁ + T₂ = 2p + 4q
    Or T₂ = 2p + 4q − T₁
    T₂ = 2p + 4q − (p + q) ⇒ p + 3q
    Hence common difference = T₂ − T₁
    = p + 3q − (p + q) = 2q

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