Arithmetic Progression

Mathematics Class Tenth

10th-Math-home


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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