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# ARITHMETIC PROGRESSION- CLASS 10

CONTENT LIST

## ARITHMETIC PROGRESSION

### Introduction

Arithmetic progression is a seriers or sequence of terms, in which difference of any two consecutive terms are constant.

The constant value is called common difference.

#### n terms of an Arithmetic progression

General form of arithmetic progression:- a, a+d, a+2d, a+3d ........ a+(n-1)d

nth term of any arithmetic progression (an) = a+(n-1)d

Terms in arithmetic progression:-

1. Two terms in AP:- a, a+d
2. Three terms in AP:- a, a+d, a-d
3. Four terms in AP:- a-3d, a-d, a+d, a+3d
4. Five terms in AP:- a-2d, a-d, a, a+d, a+2d

#### Sum of n terms of AP

Sum of n terms of an AP = n/2(a+l)

Where, a => First term
l => last term

Sum of n terms of an AP = n/2[2a + (n-1)d]

Where, a => First term
d => common difference

### ONE MARKS QUESTION OF ARITHMATIC PROGRESSION

Q1: Write first four terms of the AP, when the first term a = 10 and the common difference d = 10?

Solution:
First term (a) = 10 [Given]
Second term = a + d = 10 + 10 = 20
Third term = a + 2d = 10 + 2x10 = 30
Fourth term = a + 3d = 10 + 3x10 = 40

Q2: Find out the first term and the common difference of an AP, – 5, – 1, 3, 7, . . .

Solution: First term (a) = -5 [Given]
Common difference (d) = second term - first term = (-1) - (-5) = -1 + 5 = 4

Q3: Find out wheather given series is in AP. If it form an AP, find the common difference d and write three more terms.
3, 3 + √2, 3 + 2√2 , 3 + 3√2, . . .

Solution: Given, First term = 3, Second term = 3 + √2, Third term = 3 + 2√2, Forth term = 3 + 3√2
(Second term) - (First term) = (3 + √2) - (3) = √2
(Forth term) - (Third term) = (3 + 3√2) - (3 + 2√2) = √2
Here, we found (Second term) - (First term) = (Forth term) - (Third term) = √2
Therefore, the given series are in AP with common difference (d = √2).
So, the fifth term = a +4d = 3 + 4√2, sixth term = a + 5d = 3 + 5√2, seventh term = a + 6d = 3 + 6√2

Q4: Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .

Solution: Given, First term = 11, common difference = 8 - 11 = -3
Let -150 be the nth term of the AP

According to the question, we get
=> an = -150
=> a + (n - 1)d = -150
=> 11 + (n - 1)(-3) = -150
=> (n - 1) = (-150 - 11)/(-3) = -139/-3
Here, we are getting n - 1 as a frational number, so -150 is not the term of given AP.

Q5: Find the 25𝑡ℎ term of the A.P. −5,−5/2,0,5/2,…

Solution: Given, First term = -5, common difference = -5/2 - (-5) = -5/2 + 5 = 5/2

We know that
=> an = a + (n - 1)d
=> a25 = -5 + 24 x 5/2
=> a25 = -5 + 12 x 5
=> a25 = -5 + 60
=> a25 = 55
So, the 25th term of the AP is 55.

Q5: Find the common difference of the AP 1/𝑝,(1−𝑝)/𝑝,(1−2𝑝)/𝑝,……

Solution: Given, First term = 1-p, common difference = (1−𝑝)/𝑝 - 1/p = (1-p-1)/p = -p/p = -1

We know that
=> an = a + (n - 1)d
=> a25 = -5 + 24 x 5/2
=> a25 = -5 + 12 x 5
=> a25 = -5 + 60
=> a25 = 55
So, the 25th term of the AP is 55.

Q6: The value of 𝑎30−𝑎20 for the A.P. 2, 7, 12, 17,… is

Solution: Given, First term = 2, common difference = 7 - 2 = 5

According to the question
=> a20 = a + 19d = 2 + 19x5 = 2 + 95 = 97
=> a30 = a + 29d = 2 + 29x5 = 2 + 145 = 147

=> a30 - a20 = 147 - 97 = 50

Q7: The sum of first 20 odd natural numbers is:

Solution: Given, First term = 1, common difference = 3 - 1 = 2

We know that
=> Sn = n/2[2a + (n-1)d] = 20/2[2x1 +(20-1)2] = 10[2 + 19x2] = 10 x 40 = 400

Q8: In an AP, if d = −2, 𝑛 = 5 and 𝑎𝑛 = 0, then find the value of 𝑎.

Solution: Given, nth term (an) = 0, common difference (d) = -2, number of terms (n) = 5

We know that
=> an = a + (n-1)d
=> an = 0
=> a + (n-1)d = 0

=> a + (5-1)(-2) = 0
=> a + 4(-2) = 0
=> a - 8 = 0
=> a = 8

Q9: If the sum of first 𝑚 terms of an A.P. is 2𝑚²+3𝑚, then what it its second term?

Solution: Given, Sum of first m terms of an AP = 2𝑚²+3𝑚
According to the question
=> Sn = 2𝑚²+3𝑚
=> S1 = 2x1²+3x1 = 2 + 3 = 5
=> S2 = 2x2²+3x2 = 8 + 6 = 14

=> a1 = 5
=> a2 = 14 - 5 = 9
So, the second term of the AP is 9.

Q10: In an AP If sum of 3rd and 7th term is zero.Find 5th term?[CBSE 2022]

Solution: Let an AP be first term=a , common diffierence =d .
∴a3(third term)=a+2d
∴a7(seventh term)= a+6d
According to the question
=> a7+a3 = 0
=> (a+2d)+(a+6d)=0
=> 2a+8d = 0

=> a+4d=0
∴fifth term of AP= a+4d = 0

Q11: Determine the AP whose 3rd term is 5 and the 7th term is 9.[CBSE 2022]

Solution: We know that nth term of an AP, An= a+(n-1)d ,where a & d be the first term and common difference of an AP.
∴a3(third term)=a+2d=5 →( Eqn. 1)
∴a7(seventh term)= a+6d=9 →(Eqn.2)
Now, Euation (2) -Equation (1):- =>4d=4
=>d=1
from eqn-1 Weget,
=>a+2(1)=5 ⇒ a=3
∴ Required AP=3,4,5,6,.....

Q12: Find the sum of n terms of an AP whose nth term is given by an=5-2n .[CBSE 2022]

Solution: Given nth term of an AP, An= 5-2n
∴a1(first term)=5-2*1=3
We Know as
=>Sn=n/2(a+l)
=>Sn=n/2(3+5-2n)
=>Sn=n/2(8-2n)
=>Sn=n(4-n)
=>Sn=4n-4n^2

Q13: For what value of 𝑘 will the consecutive terms 2𝑘+1,3𝑘+3 and 5𝑘−1 form an A.P.?[CBSE 2016]

Solution: Given 2𝑘+1,3𝑘+3 and 5𝑘−1 fare an A.P.
let
2k + 1 = a1
3k + 3 = a2
5k - 1 = a3
To form an AP,
a2-a1 = a3-a2
=>2a2=a1+a3
=>2( 3k + 3)=2k + 1 + 5k - 1
=>6k + 6=7k
=>k = 6
So, k = 6

Q14: Find the 9th term from the end (towards the first term) of the A.P. 5,9,13,……,185.[CBSE 2016]

Solution:
First term(a)=5
Common difference(d) of the AP = 9 − 5 = 4 Last term of the AP = 185
We know that the nth term an AP is given by= a+(n-1)d
=>185=5+(n-1)4
=>n-1=180/4=45
=>n=46
=> 9th term from the end =(46-9)+1=38 term from starts
let us find 38 the of AP.
=>9th term from the end =(46-9)+1=38th term from starts=5+(38-1)4
=>5+37x4=153

Q15: If the common difference of an AP is 3, then what is 𝑎15−𝑎9?.

Solution:
Given,common difference (d)=3
∴𝑎15−𝑎9=(a+14d)-(a+8d)=6d=6X3=18

Q16: Find out next term of the A. P. √18,√50,√98,…

Solution:
Given A. P= √18,√50,√98,…
This be written as:
3√2 , 5√2 , 7√2, …..
First term, a = √2
d = 2√2
We know,
an= a+(n-1) d
So next term is: 4th term=3√2+3X2√2=9√2=√162