Class 10 Chapter 5 (arithmetic progression) class notes ~ THE MATH CENTRE

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

$a_{1}, a_{2}, a_{3} ....... a_{n}$ is an AP, if

$a_{1}, a_{2}, a_{3} ....... a_{n + 1} - a_{n} i s i n d e p e n d e n t o f n$
For Example: If sequence is 2, 4, 6, 8, ……

$a_{n}$ ,….. so if we take

$a_{n} = 16$ so

$a_{n + 1} = 18$ , Here

$a_{n + 1} - a_{n} = 18 - 16 = 2$ which is independent of n.

> A sequence

$a_{1}, a_{2}, a_{3} ....... a_{n}$ 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

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

Also,

$S_{n} = \frac{n}{2} [a + l]$ , where l = last term = a + (n − 1)d
For Example: (i) 50, 46, 42, … find the sum of first 10 terms
Solution:

$\begin{array}{l} G i v e n, 50, 46, 42, ..... H e r e, f i r s t t e r m (a) = 50, c o m m o n d i f f e r e n c e (d) = 46 - 50 = (- 4) \\ A n d n o o f t e r m s n = 10 \\ W e k n o w S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ \Rightarrow S_{n} = \frac{10}{2} [2 \times 50 + (10 - 1) (- 4] \\ \Rightarrow 5 [100 + 9 \times - 4] = 5 [100 - 36] \\ \Rightarrow 5 \times 64 = 320 \end{array}$ 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,

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
   350 = 17 + (n − 1)9
   350 = 17 + 9n − 9
   9n = 350 − 17 + 9 ⇒ 342
   n = 38
   We know, sum of n terms

$S_{n} = \frac{n}{2} [a + l]$

$S_{38} = \frac{38}{2} [17 + 350]$ ⇒ 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

$\frac{7 n + 1}{4 n + 27}$ . 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,

$\Rightarrow \frac{S_{n_{1}}}{S_{n_{2}}} = \frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]}$

$\Rightarrow \frac{S_{n_{1}}}{S_{n_{2}}} = \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}}$

$It is given that \frac{S_{n_{1}}}{S_{n_{2}}} = \frac{7 n + 1}{4 n + 27}$

$\Rightarrow \frac{7 n + 1}{4 n + 27} = \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} ....... (i)$

> To find ratio of the m^th terms of the two given AP’s, we replace n by (2m − 1) in equation (i). Therefore,

$\Rightarrow \frac{7 (2 m - 1) + 1}{4 (2 m - 1) + 27} = \frac{2 a_{1} + ((2 m - 1) - 1) d_{1}}{2 a_{2} + ((2 m - 1) - 1) d_{2}}$

$\Rightarrow \frac{14 m - 7 + 1}{8 m - 4 + 27} = \frac{2 a_{1} + ((2 m - 2) d_{1}}{2 a_{2} + ((2 m - 2) d_{2}}$

$\Rightarrow \frac{14 m - 6}{8 m - 23} = \frac{a_{1} + (m - 1) d_{1}}{a_{2} + (m - 1) d_{2}}$ Hence, the ratio of the mth terms of the two AP’s is

$\frac{14 m - 6}{8 m - 23}$
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