Arithmetic Progression
Mathematics Class Tenth
NCERT Exercise 5.1 Q3-4
Question: (3) For the following APs, write the first term and the common difference:
(i) `3,1,-1,-3,` .....
Solution:
Here, First term `(a_1)` = 3
We know that if differences between next term and preceding term are common, it is called common difference
Thus, `a_2-a_1=1-3=-2`
And, `a_3-a_2 = -1-1=-2`
And, `a_4-a_3=-3-(-1)=-3+1=-2`
Here, `a_(k+1)-a_k=-2` for all values of `k`
∴ Common difference `=-2`
∴ First term = 3 and common difference `=-2` Answer
Question 3(ii) `-5, -1, 3, 7, ` .....
Solution:
Given, `-5, -1, 3, 7, ` .....
Here, First term `(a_1)=-5`
We know that if differences between next term and preceding term are common, it is called common difference
Now, `a_2-a_1 = -1-(-5)=-1+5=4`
And, `a_3-a_2=3-(-1)=3-1=4`
And, `a_3-a_3 = 7-3=4`
Here, `a_(k+1)-a_k=4` for all values of `k`
∴ Common difference = 4
Thus, First Term is `-5` and commond difference is 4. Answer
Question 3. (iii) `1/3, 5/3, 9/3, 13/3,` ....
Solution:
Given, `1/3, 5/3, 9/3, 13/3,` ....
Here, First Term `(a_1) = 1/3`
We know that if differences between next term and preceding term are common, it is called common difference
Now, `a_2-a_1 = 5/3-1/3`
`=(5-1)/3=4/3`
And, `a_3-a_2=9/3-5/3`
`=(9-5)/3=4/3`
And, `a_4-a_3 = 13/3-9/3`
`=(13-9)/3=4/3`
Here, `a_(k+1)-a_k=4/3` for all values of `k`
∴ Common Difference `=4/3`
∴ First term `=1/3` and Common difference `=4/3` Answer
Question 3 (iv) ` 0.6, 1.7, 2.8, 3.9` ......
Solution:
Here, First Term `(a_1)=0.6`
We know that if differences between next term and preceding term are common, it is called common difference
Now, `a_2-a_1 = 1.7-0.6 =1.1`
And, `a_3-a2=2.8-1.7 = 1.1`
And, `a_4-a_3 = 3.9-2.8=1.1`
Here, `a_(k+1)-a_k=1.1` for all values of `k`
∴ Common difference = 1.1
Therefore, First term = 0.6 and common difference = 1.1 for the given Arithmetic Progression. Answer
Question 4. Which of the following are APs? If they form an AP, find the commond difference `d` and write three more terms.
(i) `2,4, 8, 16,` .....
Solution:
Here, `a_2-a_1 = 4-2 = 2`
And, `a_3-a_2 = 8-4 = 4`
And, `a_4-a_3 = 16-8=8`
Here, since the value of `a_(k+1)-a_k` is not equal or same for all values of `k`. Thus given list of numbers do not form an Arithmetic Progression (AP)
Question 4. (ii) `2, 5/2, 3, 7/2,` ......
Solution:
Here, `a_2-a_1 = 5/2-2`
`=(5-4)/2=1/2`
And, `a_3-a_2 = 3-5/2`
`= (6-5)/2 = 1/2`
And, `a_4-a_3 = 7/2-3`
`= (7-6)/2=1/2`
Here, value of `a_(k+1)-a_k` for all values of `k` is same thus, given list of numbers are in Arithmetic Progression
Here, First term `a_1 = 2`
And Common difference `= 1/2`
We know that, `a_n = a+(n-1)d` where `n` is number of term
∴ Fifth term
`a_5 = 2+(5-1)xx 1/2`
`= 2+ 4xx1/2 = 2+2`
`:. a_5 = 4`
Sixth term
`a_6 = 2+(6-1)xx1/2`
`= 2+5xx1/2 = (4+5)/2`
`:. a_6 = 9/2`
Seventh term
`a_7 = 2+(7-1)xx1/2`
`= 2+6xx1/2=2+3`
`:. a_7 = 5`
∴ Common difference, `d = 1/2`; and `4, 9/2,` and `5` are the three more terms for the given Arithmetic Progression (AP) Answer
Question 4. (iii) `-1.2, -3.2, -5.2, -7.2,` .......
Solution:
Given, `-1.2, -3.2, -5.2, -7.2,` .......
Here, `a_2-a_1 = -3.2-(-1.2)`
`=-3.2+2.2=-2`
And, `a_3-a_2=-5.2-(-3.2)`
`=-5.2+3.2=-2`
And, `a_4-a_3 = -7.2-(-5.2)`
`= -7.2+5.2=-2`
Here, value of `a_(k+1)-a_k` for all values of `k` is same thus, given list of numbers are in Arithmetic Progression (AP)
Here, First term `a_1 = -1.2`
And Common difference `= -2`
We know that, `a_n = a+(n-1)d` where `n` is number of term
∴ Fifth term
`a_5 = -1.2+(5-1)xx(-2)`
`= -1.2 + 4xx(-2)=-1.2-8`
`=>a_5=-9.2`
And, Sixth term
`a_6=-1.2+(6-1)xx(-2)`
`= -1.2+5xx(-2)=-1.2-10`
`:. a_6 = -11.2`
And, Seventh term
`a_7 = -1.2+(7-1)xx(-2)`
`= -1.2 +6xx(-2)= -1.2-12`
`:. a_7 = -13.2`
∴ Common difference, `d = -2`; and `-9.2, -11.2` and `-13.2` are the three more terms for the given Arithmetic Progression (AP) Answer
Question. 4 (iv) `-10, -6, -2, 2, ` .......
Solution:
Given, `-10, -6, -2, 2, ` .......
Here, `a_2-a_1 = -6-(-10)`
`=-6+10=4`
And, `a_3-a_2 = -2-(-6)`
`=-2+6 =4`
And, `a_4-a_3 = 2-(-2)`
`=2+2=4`
Here, since value of `a_(k+1)-a_k` for all values of `k` is same thus, given list of numbers are in Arithmetic Progression (AP)
Here, First term `(a_1)=-10`
And common difference = 4
We know that, `a_n = a+(n-1)d` where `n` is number of term
∴ Fifth term
`a_5= -10+ (5-1)xx4`
`=-10+4xx4=-10+16`
`:. a_5 = 6`
And Sixth term
`a_6 = -10+(6-1)xx4`
`= -10+5xx4=-10+20`
`:. a_6 = 10`
And, Seventh Term
`a_7 = -10 + (7-1)xx4`
`= -10 + 6 xx 4=-10+24`
`:. a_7 = 14`
∴ Common difference, d = 4; and 6, 10 and 14 are the three more terms for the given Arithmetic Progression (AP) Answer
Question 4. (v) ` 3, 3+sqrt2, 3+2sqrt2, 3+3sqrt2` .....
Solution:
Given, ` 3, 3+sqrt2, 3+2sqrt2, 3+3sqrt2` .....
Here, `a_2-a_1=3+sqrt3-3=sqrt2`
And, `a_3-a_2 = 3+2sqrt2-(3+sqrt2)`
`=3+2sqrt2-3-sqrt2`
`= 2sqrt2-sqrt2=sqrt2`
And, `a_4-a_3 = 3+3sqrt2-(3+2sqrt2)`
`=3+3sqrt2-3-2sqrt2`
`= 3sqrt2-2sqrt2=sqrt2`
Here, since value of `a_(k+1)-a_k` for all values of `k` is same thus, given list of numbers are in Arithmetic Progression (AP)
Here, First term `(a_1)=3`
And common difference `= sqrt2`
We know that, `a_n = a+(n-1)d` where `n` is number of term
∴ Fifth term
`a_5 = 3+(5-1)xxsqrt2`
`=3+4sqrt2`
And, Sixth term
`a_6 = = 3+(6-1)xxsqrt2`
` = 3+5sqrt2`
And, Seventh term
`a_7 = = 3+(7-1)xxsqrt2`
`=3+6sqrt2`
∴ Common difference, `d =sqrt2` ; and `3+4sqrt2, 3+5sqrt2` and `3+6sqrt2` are the three more terms for the given Arithmetic Progression (AP) Answer
Question 4. (vi) ` 0.2, 0.22, 0.222, 0.2222, ` .......
Solution:
Given, ` 0.2, 0.22, 0.222, 0.2222, ` .......
Here, `a_2-a_1 = 0.22-0.2`
`= 0.02`
And, `a_3 - a_2 = 0.222-0.22`
`= 0.002`
And, `a_4-a_3 = 0.2222-0.222`
`=0.0002`
Here, since the value of `a_(k+1)-a_k` is not equal or same for all values of `k`. Thus given list of numbers do not form an Arithmetic Progression (AP)
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