## Arithmetic Progressions

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**Class 10 NCERT Exemplar Chapter Solutions**

**Q 1(MCQ): In an AP, if d=-4,a**

_{n}=7 and a_{n}=4, then a is equal to(a) 6

(b) 7

(c) 20

(d) 28

Answer: (d) In an AP, a

_{n}=a+(n-1)d

⇒ 4=a+(7-1) (-4)

⇒ 4=a+6(-4)

⇒ 4+24=a

∴ a=28

**Q 2(mcq): In an AP, if a=3.5,d=0 and n=101, then a**

_{n}will be(a) 0

(b) 3.5

(c) 103.5

(d) 104.5

Answer: (b) 3.5

For an AP, a

_{n}=a+(n-1)d=3.5+(101-1)×0

∴ =3.5

**Q 3(mcq): The list of numbers are -10,-6,-2,2,.... is**

(a) an AP with d=-16

(b) an AP with d=4

(c) an AP with d=-4

(d) not an AP

Answer: (b) The given numbers are -10,-6,-2,2...is

Here, a

_{1}= -10,

a

_{2 }= -6,

a

_{3 }= -2

and a

_{4 }=2...

Since, a

_{2 }- a

_{1 }= -6-(-10)

= -6+10=4

a

_{3}-a

_{2}= -2-(-6)

=-2+6=4

a

_{4}-a

_{3}=2-(-2)

=2+2=4

Each successive term of given list has the same difference i.e.,4.

So, the given list forms an AP with common difference, d=4.

**Q 4(mcq): The first four terms of an AP whose first term is -2 and the common difference is -2 are**

(a) -2,0,2,4

(b) -2,4,-8,16

(c) -2,-4,-6,-8

(d) -2,-4,-8,-16

Answer: (c) -2,-4,-6,-8

Let the first four terms of an AP are a,a+d,a+2d and a+3d.

Given, that first term, a=-2 and common difference, d=-2, then we have an AP as follows

-2,-2,-2,-2+2(-2),-2+3(-2)

= -2,-4,-6,-8

**Q 5(mcq): The 21st term of an AP whose first two terms are -3 and 4 is**

(a) 17

(b) 137

(c) 143

(d) -143

Answer: (b) 137

The first two terms of an AP are a=-3 and a+d=4.

⇒ -3+d=4

Common difference, d=7

∴ a

_{21}=a+(21-1) d [∵ a

_{n}=a+(n-1) d]

=-3+(20)7

=-3+140=137

**Q 6: Which of the following form of an AP? Justify your answer.**

(i) -1,-1,-1,-1....

(ii) 0,2,0,2

(iii) 1,1,2,2,3,3,...

(iv) 11,22,33

Answer: (i) Here, t

_{1}=1,t

_{2}=-1, t

_{3}=-1 and t

_{4}=-1

t

_{2 }- t

_{1 }=-1+1=0

t

_{3 }- t

_{2 }=-1+1=0

t

_{4 }- t

_{3 }=-1+1=0

Obviously, the difference of successive terms is same, therefore given list of numbers form an AP.

(ii) Here, t

_{1}=0,t

_{2}=2,t

_{3}=0 and t

_{4}2

t

_{2 }- t

_{1 }= 2 - 0 = 2

t

_{3 }- t

_{2 }= 0 - 2 = -2

t

_{4 }- t

_{3 }= 2 - 0 = 2

Since the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iii) Here, t

_{1}, t

_{2}=1,t

_{3}=2 and t

_{4}=2

t

_{2}-t

_{1}=1-1=0

t

_{3}-t

_{2}=2-1=1

t

_{4}-t

_{2}=2-2=0

Since the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iv) Here, t

_{1}=11,t

_{2}=22 and t

_{3}=33

t

_{2 }- t

_{1 }= 22-11=11

t

_{3 }- t

_{2 }= 33-22=11

t

_{4}-t

_{3 }= 33-22=11

Since the difference of successive terms is same, therefore given list of numbers form an AP.

**Q 7: For the AP -3,-7,-11,....can we find directly a**

_{30}-a_{20}without actually finding a_{30}and a_{20}? Give reason for your answer.Answer: TRUE (✔)

∵ n the term of an AP, a

_{n}= a(n-1)d

∴ a

_{30}=a+(30-1) d=a+29d

and a

_{20}=a+(20-1) d=a+19d

Now,a

_{30}-a

_{20}=(a+29d)-(a+19d)=10d

and from given AP common difference, d=-7-(-3)=-7+3

= -4

∴ a

_{30}-a

_{20}=10(-4)=-40

**Q 8: Two AP's have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms? Why?**

Answer: Let the same common difference of two AP's =

*d*.

Given that, the first term of first AP and second AP are 2 and 7 respectively, then the AP's are

2, 2+d, 2+2d, 2+3d....

and 7,7+d,7+2d,7+3d,...

Now, 10th terms of first and second AP's are 2+9 d and 7+9 d, respectively.

So, their differences is 7+9d-(2+9d))=5

Also, 21st terms of first and second AP's are 2+20d and 7+20d, respectively.

So, their difference is 7+20d-(2+9d)=5

Also, if the a

_{n}and b

_{n}are the nth terms of first second AP.

Then, b

_{n}-a

_{n}=[7+(n-1)d]-[2+(n-1)d]=5

Hence, the difference between any two corresponding terms of such AP's is the same as the difference between their first terms.

**Q 9: The taxi fare after each km, when the fare is ₹15 for the first km and ₹8 for each additional km, does not form an AP as the total fare (in₹) after each km is 15,8,8,8,... . Is the statement true? Give reasons.**

Answer: no, because the total fare (in₹) after each km is

15,(15+8), (15+2×8),(15+3×8),...=15,23,31,39,...

Let t

_{1}=15,t

_{2}=23,t

_{3}=31 and t

_{4}=39

Now, t

_{2}-t

_{1}=23-15=8

t

_{3}-t

_{2}=31-23=8

t

_{4}-t

_{3}=39-31=8

Since, all the successive terms of the given list have same difference i.e., common difference=8

Hence, the total fare after each km form an AP.

**Q 10: Justify whether it is true to say that the following are the nth terms of an AP.**

(i) 2n-3

(ii) 3n

^{2}+5

(iii) 1+n+n

^{2}

Answer: (i) Yes, here a

_{n}=2n - 3

Putting n=1, a

_{1}=2(1)-3=-1

Putting n=2, a

_{2}=2(2)-3=1

Putting n=3, a

_{3}=2(3)-3=3

Putting n=4, a

_{4}=2(4)-3=5

List of numbers becomes-1,1,3,...

Here, a

_{2}-a

_{1}=1-(-1)=1+1=2

a

_{3}-a

_{2}=3-1=2

a

_{4}-a

_{3}=5-3=2

∵ a

_{2}-a

_{1}=a

_{3}-a

_{2}=a

_{4}-a

_{3}=...

Hence, 2n-3 is the nth term of an AP.

(ii) No,

Since a

_{n}=3n

^{2}+5

Put n=1, a

_{1}=3(1)

^{2}+5=8

Put n=2, a

_{2}=3(2)

^{2}+5=3(4)+5=17

Put n=3, a

_{3}=3(3)

^{2}+5=3(9)+5+5=27+5=32

∴ the list of number becomes 8,17,32,...

Here, a

_{2}-a

_{1}=17-8=9

a

_{3}-a

_{2}=32-17=15

∴ a

_{2}-a

_{1}≠ a

_{3}-a

_{2}

Since, the successive difference of the list is not same. So, it does not form an AP.

(iii) No, here a

_{n}=1+n+n

^{2}

For n=1, a

_{1}=1+1+(1)

^{2}=3

For n=2, a

_{2}=1+2+(2)

^{2}=1+2+4=7

For n=3, a

_{3}=1+3+(3)

^{2}=1+3+9=13

So, the list of number becomes 3,7,13,...

Here, a

_{2}-a

_{1}= 7 - 3 = 4

a

_{3}-a

_{2 }= 13 - 7 = 6

∴ a

_{2}-a

_{1}≠ a

_{3}-a

_{2}

Since, the successive difference of the list is not same. So, it does not form an AP.

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