Arithmetic Progression
Mathematics Class Tenth
NCERT Exercise 5.1 Q4(viii-xv)
Question. 4 (vii) `0, -4, -8, -12` ......
Solution:
Given, `0, -4, -8, -12` ......
Here, `a_2-a_1 = -4-0=-4`
And, `a_3-a_2 = -8-(-4)`
`= -8+4=-4`
And, `a_4-a_3 = -12-(-8)`
`= -12+8=-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 `0`
And common difference `= -4`
We know that, `a_n = a+(n-1)d` where `n` is number of term
∴ Fifth term
`a_5 = 0+(5-1)xx(-4)`
`= 4xx(-4)=-16`
And Sixth term
`a_6 = 0+(6-1)xx(-4)`
`= 5xx(-4)=-20`
`And, Seventh term
`a_7 = 0+(7-1)xx(-4)`
`= 6xx(-4) = -24`
∴ Common difference, `d = -4`; and `-16, -20` and `-24` are the three more terms for the given Arithmetic Progression (AP) Answer
Question . 4 (viii) `-1/2, -1/2, -1/2, -1/2` .....
Solution:
Given, `-1/2, -1/2, -1/2, -1/2` .....
Here, `a_2-a_1=a_3-a_2=a_4-a_3 =0`
i.e. commond difference =0
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)
Since, common difference = 0
∴ Fifth term `(a_5) = -1/2`
And, Sixth term, `(a_6)=-1/2`
And, Seventh term, `(a_7)=-1/2`
∴ Common difference, d = 0; and `-1/2, -1/2` and `-1/2` are the three more terms for the given Arithmetic Progression (AP) Answer
Question. 4 (ix) `1, 3, 9, 27, ` .......
Solution:
Given, `1,3,9,27` ......
Here, `a_2-a_1 = 3-1=2`
And, `a_3-a_2 = 9-3=6`
And, `a_4-a_3 = 27-9=18`
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 (x) `a, 2a, 3a, 4a,` .......
Solution:
Given, `a, 2a, 3a, 4a,` .......
Here, `a_2-a_1 = 2a-a = a`
And, `a_3-a_2 = 3a-2a=a`
And, `a_4-a_3 = 4a-3a=a`
Here, since value of `a_(k+1)-a_k` for all values of `k` is same thus, given list of numbers is in Arithmetic Progression (AP)
∴ First term `=a`
And common difference `=a`
We know that, `a_n = a+(n-1)d` where `n` is number of term
∴ Fifth term
`a_5 = a+(5-1)xxa`
`=>a_5 = a+4a=5a`
And Sixth term
`a_6 = a+(6-1)xxa`
`a_6 = a+5a=6a`
And, Seventh term
`a_7 = a+(7-1)xxa`
`a_7 = a+6a=7a`
∴ Common difference, `d = a`; and `5a, 6a` and `7a` are the three more terms for the given Arithmetic Progression (AP) Answer
Question. 4 (xi) `a, a^2, a^3, a^4,` .......
Solution:
Given, `a, a^2, a^3, a^4,` .......
Here, `a_2-a_1 = a^2-a`
`=a(a-1)`
And, `a_3-a_2 = a^3-a^2`
`=a^2(a^2-1)`
And, `a_4-a_3 = a^4-a^3`
`=a^3(a^2-1)`
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 (xii) `sqrt2, sqrt8, sqrt18, sqrt32` ....
Solution:
Given, `sqrt2, sqrt8, sqrt18, sqrt32` ....
Here, `a_2-a_1 = sqrt8-sqrt2`
`=sqrt(4xx2)-sqrt2`
`=2sqrt2-sqrt2=sqrt2`
And, `a_3-a_2 = sqrt18-sqrt8`
`=sqrt(9xx2)-sqrt(4xx2)`
`=3sqrt2-2sqrt2 =sqrt2`
And, `a_4-a_3=sqrt32-sqrt18`
`=sqrt(16xx2)-sqrt(9xx2)`
`=4sqrt2-3sqrt2=sqrt2`
Here, since value of `a_(k+1)-a_k` for all values of `k` is same thus, given list of numbers is in Arithmetic Progression (AP)
Here, first term `= sqrt2`
And common difference `=sqrt2`
We know that, `a_n = a+(n-1)d` where `n` is number of term
∴ Fifth term
`a_5 = sqrt2+(5-1)xxsqrt2`
`=sqrt2+4sqrt2`
`=>a_5=5sqrt2`
By squaring both sides, we get
`(a_5)^2=(5sqrt2)^2`
`=>(a_5)^2=25xx2=50`
`=>a_5=sqrt50`
And, Sixth term
`a_6=sqrt2+(6-1)xxsqrt2`
`=sqrt2+5sqrt2`
`=>a_6=6sqrt2`
By squaring both sides, we get
`(a_6)^2=(6sqrt2)^2`
`=>(a_6)^2=36xx2=72`
`=>a_6 = sqrt72`
And Seventh term
`a_7=sqrt2+(7-1)xxsqrt2`
`=sqrt2+6sqrt2`
`=>a_7=7sqrt2`
By squaring both sides, we get
`(a_7)^2=(7sqrt2)^2`
`=>(a_7)^2=49xx2=98`
`=>a_7 = sqrt98`
∴ Common difference, `d = sqrt2`; and `sqrt50, sqrt72` and `sqrt98` are the three more terms for the given Arithmetic Progression (AP) Answer
Question. 4 (xiii) `sqrt3, sqrt 6, sqrt9, sqrt12` ........
Solution:
Given, `sqrt3, sqrt 6, sqrt9, sqrt12` ........
Here, `a_2-a_1=sqrt6-sqrt3`
And, `a_3-a_2=sqrt9-sqrt6`
`=3-sqrt6`
And, `a_4-a_3 = sqrt12-sqrt9`
`=sqrt(4xx3)-3`
`=2sqrt3-3`
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 (xiv) `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`
`=9-1=8`
And, `a_3-a_2=5^2-3^2`
`=25-9=16`
And, `a_4-a_3=7^2-5^2`
`=49-25=24`
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. (xv) `1^2, 5^2, 7^2, 73` .......
Solution:
Given, `1^2, 5^2, 7^2, 73` .......
Here, `a_2-a_1=5^2-1^2`
`=25-1=24`
And, `a_3-a_2=7^2-5^2`
`=49-25=24`
And, `a_4-a_3=73-7^2`
`=73-49=24`
Here, since value of `a_(k+1)-a_k` for all values of `k` is same thus, given list of numbers is in Arithmetic Progression (AP)
We know that, `a_n = a+(n-1)d` where `n` is number of term
∴ Fifth term
`a_5=1^2+(5-1)xx24`
`=1+4xx24=1+96`
`=>a_5=97`
And, Sixth term
`a_6=1^2+(6-1)xx24`
`=1+5xx24=1+120`
`=>a_6=121`
And, Seventh term
`a_7=1^2+(7-1)xx24`
`=1+6xx24=1+144`
`a_7=145`
∴ Common difference, d = 24; and 97, 121 and 144 are the three more terms for the given Arithmetic Progression (AP) Answer
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