CHAPTER-5 ARITHMETIC PROGRESSIONS

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

Explanation:

The given circumstances can be expressed as

 Taxi’s fare for 1 km is =15

 Taxi’s fare for first 2 kms =15+8=23

 Taxi’s fare for first 3 kms=23+8=31

 Taxi’s fare for first 4 kms=31+8=39

Hence 15,23,31,39 will form an A.P.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Explanation:

A cylinder contains 1/4 litre of air.

The hoover pump simultaneously removes 1/4 of the cylinder's residual air with each stroke.

Hence, quantities will be

Here, we can see that there is no difference between the terms in this series that are adjacent to one another.

This is not an AP series.

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.

Explanation:

The given condition can be expressed as;

First metre of a well's excavation costs Rs. 150

150 plus 50 equals $200 for the first two metres of a well's excavation.

The first 3 metres of a well's excavation cost Rs. 200 plus Rs. 50, or Rs. 250.

Digging a well for the first 4 metres will cost you.Rs. 250 plus Rs. 50, or Rs. 300.

so forth.

It is evident that the numbers 150, 200, 250, and 300... create an A.P. with a shared difference of 50 between each term.

(iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound interest at 8% per annum.

Explanation:

The amount will be after each year if we deposited Rs. P. at r% compound interest every year for n years.

Hence after each year the money will be

There is nothing about them that is similar.

There is no A.P. here.

Explanation:

(i) a=10 ,d=10

arithmetic progression series be a1,a2,a3,a4,a5….

a1 = a = 10

a2 = a1+d = 10+10 = 20

a3 = a2+d = 20+10 = 30

a4 = a3+d = 30+10 = 40

a5 = a4+d = 40+10 = 50

A.P series achieved 10,20,30,40,.....

(ii) a = – 2, d = 0

arithmetic progression will be a1,a2,a3,a4,a5…..

a1 = a = -2

a2 = a1+d = – 2+0 = – 2

a3 = a2+d = – 2+0 = – 2

a4 = a3+d = – 2+0 = – 2

A.P series achieved -2,-2,-2,-2….

(iii) a = 4, d = – 3

arithmetic progression series will be a1,a2,a3,a4,a5….

a1 = a = 4

a2 = a1+d = 4-3 = 1

a3 = a2+d = 1-3 = – 2

a4 = a3+d = -2-3 = – 5

A.P series achieved 4,1,-2,-5..

(iv) a = – 1, d = ½

arithmetic progression series achieved a1,a2,a3,a4..

a2 = a1+d = -1+1/2 = -1/2

a3 = a2+d = -1/2+1/2 = 0

a4 = a3+d = 0+1/2 = 1/2

The A.P series achieved -1,-½,0 and ½

(v) a = – 1.25, d = – 0.25

arithmetic progression series will be a1,a2,a3,a4,a5,....

a1 = a = – 1.25

a2 = a1 + d = – 1.25-0.25 = – 1.50

a3 = a2 + d = – 1.50-0.25 = – 1.75

a4 = a3 + d = – 1.75-0.25 = – 2.00

A.P series achieved - 1.25,-1.50,- 2.00

Explanation:

(i) In the presented series

3,1,-1,-3…

First term is a=3

As common difference d=second term-first term

1-3=2

d=-2

(ii)In the presented series 

-5,-1,3,7

First terma=-5

As Common difference d=second term-first term

(-1)-(-5)=-1+5=4

(iii)In the presented series 

1/3, 5/3, 9/3, 13/3..

First term a=⅓

As Common difference d=second term-first term

5/3-⅓=4/3

(iv)In the presented series 

0.6,1.7,2.8,3.9

First term a=0.6

As Common difference d=second term-first term

1.7-0.6=1.1

Explanation:

(i) Presented us

common difference is

a2 – a1 = 4 – 2 = 2

a3 – a2 = 8 – 4 = 4

a4 – a3 = 16 – 8 = 8

an+1 – an

series is not forming an A.P

(ii)Presented 2, 5/2.

common difference is

a2 – a1 = 4 – 2 = 2

a3 – a2 = 8 – 4 = 4

a4 – a3 = 16 – 8 = 8

Therefore, an+1 – an or the common difference is not the same.

Thus presented series are not forming an A.P.

Consecutive three terms are

a5 = 7/2+1/2 = 4

a6 = 4 +1/2 = 9/2

a7 = 9/2 +1/2 = 5

(iii) Presented -1.2, – 3.2, -5.2, -7.2 

Here

a2 – a1 = (-3.2)-(-1.2) = -2

a3 – a2 = (-5.2)-(-3.2) = -2

a4 – a3 = (-7.2)-(-5.2) = -2

Hence an+1 – an or common difference is the same every time.

Thus, d = -2 and the given series are in A.P.

The consecutive  three terms are;

a5 = – 7.2-2 = -9.2

a6 = – 9.2-2 = – 11.2

a7 = – 11.2-2 = – 13.2

(iv) Presented, -10, – 6, – 2, 2 …

terms and their difference are

a2 – a1 = (-6)-(-10) = 4

a3 – a2 = (-2)-(-6) = 4

a4 – a3 = (2 -(-2) = 4

Since, an+1 – an or the common difference is same every time.

Thus, d = 4 and the given numbers are in A.P.

The next three terms are:

a5 = 2+4 = 6

a6 = 6+4 = 10

a7 = 10+4 = 14

(v) Presented, 3, 3+√2, 3+2√2, 3+3√2

Here

a2 – a1 = 3+√2-3 = √2

a3 – a2 = (3+2√2)-(3+√2) = √2

a4 – a3 = (3+3√2) – (3+2√2) = √2

Hence, an+1 – an or the common difference is the same every time.

Thus, d = √2 and the given series forms an A.P.

The Next three terms are;

a5 = (3+√2) +√2 = 3+4√2

a6 = (3+4√2)+√2 = 3+5√2

a7 = (3+5√2)+√2 = 3+6√2

(vi) Presented 0.2, 0.22, 0.222, 0.2222 …

a2 – a1 = 0.22-0.2 = 0.02

a3 – a2 = 0.222-0.22 = 0.002

a4 – a3 = 0.2222-0.222 = 0.0002

Hence, an+1 – an or the common difference is not the same every time.

The Presented series doesn’t form an A.P.

(vii) Presented  0, -4, -8, -12 …

a2 – a1 = (-4)-0 = -4

a3 – a2 = (-8)-(-4) = -4

a4 – a3 = (-12)-(-8) = -4

 an+1 – an or the common difference similar every time.

Therefore d = -4 and the given series forms an A.P.

The, consecutive three terms are

a5 = -12-4 = -16

a6 = -16-4 = -20

a7 = -20-4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….

a2 – a1 = (-1/2) – (-1/2) = 0

a3 – a2 = (-1/2) – (-1/2) = 0

a4 – a3 = (-1/2) – (-1/2) = 0

 an+1 – an or the common difference is the similar every time.

Thus, d = 0 and the given series forms an A.P.

The, consecutive three terms are

a5 = (-1/2)-0 = -1/2

a6 = (-1/2)-0 = -1/2

a7 = (-1/2)-0 = -½

(ix) 1, 3, 9, 27 …

a2 – a1 = 3-1 = 2

a3 – a2 = 9-3 = 6

a4 – a3 = 27-9 = 18

 an+1 – an or the common difference is not similar every time.

Thus, the given series doesn’t form an A.P.

(x) a, 2a, 3a, 4a …

a2 – a1 = 2a–a = a

a3 – a2 = 3a-2a = a

a4 – a3 = 4a-3a = a

 an+1 – an or the common difference is similar  every time.

Hence, d = a and the given series forms an A.P.

The consecutive three terms are;

a5 = 4a+a = 5a

a6 = 5a+a = 6a

a7 = 6a+a = 7a

(xi) a, a2, a3, a4 

a2 – a1 = a2–a = a(a-1)

a3 – a2 = a3 – a2 = a2(a-1)

a4 – a3 = a4 – a3 = a3(a-1)

 an+1 – an or the common difference is not similar every time.

 the given series will not  form an A.P.

(xii) √2, √8, √18, √32 …

a2 – a1 = √8-√2  = 2√2-√2 = √2

a3 – a2 = √18-√8 = 3√2-2√2 = √2

a4 – a3 = 4√2-3√2 = √2

 an+1 – an or the common difference is similar every time.

Thus, d = √2 and the given series forms an A.P.

The consecutive three terms are;

a5 = √32+√2 = 4√2+√2 = 5√2 = √50

a6 = 5√2+√2 = 6√2 = √72

a7 = 6√2+√2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …

a2 – a1 = √6-√3 = √3×√2-√3 = √3(√2-1)

a3 – a2 = √9-√6 = 3-√6 = √3(√3-√2)

a4 – a3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)

an+1 – an or the common difference is not similar every time.

the given series won't  form an A.P.

(xiv) 12, 32, 52, 72 …

Or, 1, 9, 25, 49 …..

a2 − a1 = 9−1 = 8

a3 − a2 = 25−9 = 16

a4 − a3 = 49−25 = 24

 an+1 – an or the common difference is not similar every time.

The given series won't  form an A.P.

(xv) 12, 52, 72, 73 …

Or 1, 25, 49, 73 …

Here

a2 − a1 = 25−1 = 24

a3 − a2 = 49−25 = 24

a4 − a3 = 73−49 = 24

an+1 – an or the common difference is similar every time.

Thus d = 24 and the presented series forms an A.P.

The consecutive  three terms are;

a5 = 73+24 = 97

a6 = 97+24 = 121

a7 = 121+24 = 145

Explanation:

(i)In, First term,

 a = 7

Common difference, d = 3

Number of terms, n = 8,

to find the nth term, an = ?

 we notice that  for an A.P.,

an = a+(n−1)d

keeping the values,

=> 7+(8 −1) 3

=> 7+(7) 3

=> 7+21 = 28

Thus, an = 28

(ii) In First term, a = -18

Common difference, d = ?

Number of terms, n = 10

Nth term, an = 0

As we notice, for an A.P.,

an = a+(n−1)d

Keeping the values,

0 = − 18 +(10−1)d

18 = 9d

d = 18/9 = 2

Thus, common difference, d = 2

(iii) In presented First term, a = ?

Common difference, d = -3

Number of terms, n = 18

Nth term, an = -5

As we notice  that  for an A.P.,

an = a+(n−1)d

keeping the values,

−5 = a+(18−1) (−3)

−5 = a+(17) (−3)

−5 = a−51

a = 51−5 = 46

, a = 46

(iv)In  presented First term, a = -18.9

Common difference, d = 2.5

Number of terms, n = ?

Nth term, an = 3.6

we notice, for an A.P.,

an = a +(n −1)d

Keeping the values,

3.6 = − 18.9+(n −1)2.5

3.6 + 18.9 = (n−1)2.5

22.5 = (n−1)2.5

(n – 1) = 22.5/2.5

n – 1 = 9

n = 10

n = 10

(v) In presented First term, a = 3.5

Common difference, d = 0

Number of terms, n = 105

Nth term, an = ?

We notice, for an A.P.,

an = a+(n −1)d

Keeping the values,

an = 3.5+(105−1) 0

an = 3.5+104×0

an = 3.5

an = 3.5

Explanation:

(i) Presented that

A.P. = 10, 7, 4, …

Thus, we can achieve,

First term, a = 10

Common difference, d = a2 − a1 = 7−10 = −3

As we noticed, for an A.P.,

an = a +(n−1)d

Keeping the values here

a30 = 10+(30−1)(−3)

a30 = 10+(29)(−3)

a30 = 10−87 = −77

Correct answer is option C.

(ii) Presented that ,

A.P. = -3, -1/2, ,2 …

 we can find,

First term a = – 3

Common difference, d = a2 − a1 = (-1/2) -(-3)

⇒(-1/2) + 3 = 5/2

we know, for an A.P.,

an = a+(n−1)d

Keeping the values here

a11 = -3+(11-1)(5/2)

a11 = -3+(10)(5/2)

a11 = -3+25

a11 = 22

the answer is option B.

Explanation:

(i) In the presented A.P., 2,2 , 26

1st and 3rd terms are,

a = 2

a3 = 26

we notice, for an A.P.,

an = a+(n −1)d

Keeping the values here,

a3 = 2+(3-1)d

26 = 2+2d

24 = 2d

d = 12

a2 = 2+(2-1)12

= 14

14 is the missing term here.

(ii) In the given A.P., , 13, ,3

a2 = 13 and

a4 = 3

 We notice, for an A.P.,

an = a+(n−1) d

Keeping the values here,

a2 = a +(2-1)d

13 = a+d ………………. (i)

a4 = a+(4-1)d

3 = a+3d ………….. (ii)

On removing equation (i) from (ii), we get,

– 10 = 2d

d = – 5

From equation (i), keeping the value of d, we get

13 = a+(-5)

a = 18

a3 = 18+(3-1)(-5)

= 18+2(-5) = 18-10 = 8

Thus, the missing terms are 18 and 8

(iii) In the presented  A.P.,

a = 5 and

a4 = 19/2

 we notice, for an A.P.,

an = a+(n−1)d

 keeping the values here,

a4 = a+(4-1)d

19/2 = 5+3d

(19/2) – 5 = 3d

3d = 9/2

d = 3/2

a2 = a+(2-1)d

a2 = 5+3/2

a2 = 13/2

a3 = a+(3-1)d

a3 = 5+2×3/2

a3 = 8

Thus, the missing terms are 13/2 and 8

(iv) In the presented A.P.,

a = −4 and

a6 = 6

we notice, for an A.P.,

an = a +(n−1) d

Here keeping the values here,

a6 = a+(6−1)d

6 = − 4+5d

10 = 5d

d = 2

a2 = a+d = − 4+2 = −2

a3 = a+2d = − 4+2(2) = 0

a4 = a+3d = − 4+ 3(2) = 2

a5 = a+4d = − 4+4(2) = 4

thus the missing terms are −2, 0, 2, and 4 

(v) In the presented  A.P.,

a2 = 38

a6 = −22

we notice, for an A.P.,

an = a+(n −1)d

Here  keeping  the values here,

a2 = a+(2−1)d

38 = a+d ……………………. (i)

a6 = a+(6−1)d

−22 = a+5d …………………. (ii)

On removing equation (i) from (ii), we get

− 22 − 38 = 4d

−60 = 4d

d = −15

a = a2 − d = 38 − (−15) = 53

a3 = a + 2d = 53 + 2 (−15) = 23

a4 = a + 3d = 53 + 3 (−15) = 8

a5 = a + 4d = 53 + 4 (−15) = −7

Thus, the missing terms are 53, 23, 8, and −7,

Explanation:

From  A.P. series as3, 8, 13, 18, …

First term, a = 3

Thus the Common difference, d = a2 − a1 = 8 − 3 = 5

nth term of given A.P. be 78. Now as we know,

an = a+(n−1)d

Hence

78 = 3+(n −1)5

75 = (n−1)5

(n−1) = 15

n = 16

Thus the 16th term of this A.P. is 78.

Explanation:

(i)In presented, 7, 13, 19, …, 205 is the A.P

First term, a = 7

TheCommon difference, d = a2 − a1 = 13 − 7 = 6

A.P. has terms

an = 205

we notice, for an A.P.,

an = a + (n − 1) d

Thus, 205 = 7 + (n − 1) 6

198 = (n − 1) 6

33 = (n − 1)

n = 34

In  this presented series has 34 terms in it.

First term, a = 18

Common difference, d = a2-a1 =

d = (31-36)/2 = -5/2

no terms in this A.P.

an = -47

we notice, for an A.P.,

an = a+(n−1)d

-47 = 18+(n-1)(-5/2)

-47-18 = (n-1)(-5/2)

-65 = (n-1)(-5/2)

(n-1) = -130/-5

(n-1) = 26

n = 27

Thus, this given A.P. has 27 terms in it

Explanation:

In the presented series, A.P. 11, 8, 5, 2..

The First term, a = 11

Common difference, d = a2−a1 = 8−11 = −3

Consider −150 be the nth term of this A.P.

We notice, for an A.P.,

an = a+(n−1)d

-150 = 11+(n -1)(-3)

-150 = 11-3n +3

-164 = -3n

n = 164/3

n is not an integer but a fraction.

Thus, – 150 is not a term of this A.P.

Explanation:

Presented that,

11th term, a11 = 38

and 16th term, a16 = 73

As We notice that,

an = a+(n−1)d

a11 = a+(11−1)d

38 = a+10d ………………………………. (i)

In the similar  way,

a16 = a +(16−1)d

73 = a+15d ………………………………………… (ii)

On removing equation (i) from (ii), we get

35 = 5d

d = 7

by equation (i), we can give,

38 = a+10×(7)

38 − 70 = a

a = −32

a31 = a +(31−1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

Thus, 31st term is 178

Explanation:

Presented that,

3rd term, a3 = 12

50th term, a50 = 106

As We notice that,

an = a+(n−1)d

a3 = a+(3−1)d

12 = a+2d ……………………………. (i)

In the similar way,

a50 = a+(50−1)d

106 = a+49d …………………………. (ii)

removing equation (i) from (ii), we get

94 = 47d

d = 2 = common difference

By equation (i), we can give,

12 = a+2(2)

a = 12−4 = 8

a29 = a+(29−1) d

a29 = 8+(28)2

a29 = 8+56 = 64

Thus,, 29th term is 64

Explanation:

Presented that,

3rd term, a3 = 4

and 9th term, a9 = −8

We notice  that,

an = a+(n−1)d

Thus

a3 = a+(3−1)d

4 = a+2d ……………………………………… (i)

a9 = a+(9−1)d

−8 = a+8d ………………………………………………… (ii)

On removing equation (i) from (ii), we present here,

−12 = 6d

d = −2

By equation (i), we can derive,

4 = a+2(−2)

4 = a−4

a = 8

Consider the nth term of this A.P. be zero.

an = a+(n−1)d

0 = 8+(n−1)(−2)

0 = 8−2n+2

2n = 10

n = 5

Thus the 5th term of this A.P. is 0.

Explanation:

We notice that, for an A.P series;

an = a+(n−1)d

a17 = a+(17−1)d

a17 = a +16d

In the similar way,

a10 = a+9d

Presented in the question,

a17 − a10 = 7

(a +16d)−(a+9d) = 7

7d = 7

d = 1

Thus common difference is 1.

Explanation:

Presented A.P. is 3, 15, 27, 39, …

first term, a = 3

Common difference, d = a2 − a1 = 15 − 3 = 12

We notice that,

an = a+(n−1)d

Thus,

a54 = a+(54−1)d

⇒3+(53)(12)

⇒3+636 = 639

a54 = 639+132=771

Notice the term of this A.P. which is 132 greater than a54, i.e.771.

Consider nth term be 771.

an = a+(n−1)d

771 = 3+(n −1)12

768 = (n−1)12

(n −1) = 64

n = 65

Thus,, the 65th term was 132 is greater than the 54th term.

Explanation:

First term of two APs be a1 and a2 considerably,

Common difference between these APs is d.

For the first A.P.,we notice,

an = a+(n−1)d

Thus,

a100 = a1+(100−1)d

= a1 + 99d

a1000 = a1+(1000−1)d

a1000 = a1+999d

For second A.P., 

an = a+(n−1)d

Therefore,

a100 = a2+(100−1)d

= a2+99d

a1000 = a2+(1000−1)d

= a2+999d

Thus difference among 100th term of the two APs = 100

Thus, (a1+99d) − (a2+99d) = 100

a1−a2 = 100……………………………………………………………….. (i)

Difference among 1000th terms of the two APs

(a1+999d) − (a2+999d) = a1−a2

By equation (i),

difference, a1−a2 = 100

Thus, the difference among 1000th terms of the two A.P. will be 100.

Explanation:

The three-digit numbers that are divisible by 7 are;

First number = 105

Second number = 105+7 = 112

Third number = 112+7 =119

105, 112, 119, …

All three-digit numbers are divisible by 7, and they are all terms of an A.P. with 105 as the first term and 7 as the most frequent difference.

999 is the greatest three-digit number that can exist.

The remaining after dividing 999 by 7 is 5.

Hence, 999-5 = 994 is the highest three-digit number that may be divided by 7.

Here is the list of episodes.

105, 112, 119, …, 994

As 994 is the  nth term of this A.P.

first term, a = 105

Thus common difference, d = 7

an = 994

n = ?

We notice,

an = a+(n−1)d

994 = 105+(n−1)7

889 = (n−1)7

(n−1) = 127

n = 128

Thus, 128 three-digit numbers will be  divisible by 7.

Explanation:

4 having the first multiple that is greater than 10 is 12.

Consecutive  multiple will be 16.

The series achieved  as;

12, 16, 20, 24, …

These are divisible by 4 and , all these are terms of an A.P. with the first term as 12 and common difference as 4.

We divide 250 by 4, the remainder is  2. Thus, 250 − 2 = 248 is divisible by 4.

The series presented as:

12, 16, 20, 24, …, 248

Thus 248 be the nth term of this A.P.

First term, a = 12

Common difference, d = 4

an = 248

an = a+(n−1)d

248 = 12+(n-1)×4

236/4 = n-1

59  = n-1

n = 60

Thus , 4 has 60 multiples between 10 and 250

Explanation:

In both APs as; 63, 65, 67,… and 3, 10, 17,….

Considering first AP,

63, 65, 67, …

First term, a = 63

Common difference, d = a2−a1 = 65−63 = 2

 nth term of this A.P. = an = a+(n−1)d

an= 63+(n−1)2 = 63+2n−2

an = 61+2n ………………………………………. (i)

Considering second AP,

3, 10, 17, …

First term, a = 3

Common difference, d = a2 − a1 = 10 − 3 = 7

Thus,

nth term of this A.P. = 3+(n−1)7

an = 3+7n−7

an = 7n−4 ……………………………………………………….. (ii)

Thus, nth term of these A.P.s are equal to each other.

considering both these equations, we achieve,

61+2n = 7n−4

61+4 = 5n

5n = 65

n = 13

Thus 13th terms of both these A.P.s are equal to each other

Explanation:

Third term, a3 = 16

We notice

a +(3−1)d = 16

a+2d = 16 ………………………………………. (i)

7th term increasess the 5th term by 12.

a7 − a5 = 12

[a+(7−1)d]−[a +(5−1)d]= 12

(a+6d)−(a+4d) = 12

2d = 12

d = 6

By equation (i), we get,

a+2(6) = 16

a+12 = 16

a = 4

Thus, A.P. will be4, 10, 16, 22, …

Explanation:

A.P. is3, 8, 13, …, 253

Common difference, d= 5.

We can present the given AP in reverse order as;

253, 248, 243, …, 13, 8, 5

New AP,

First term, a = 253

Common difference, d = 248 − 253 = −5

n = 20

Thus, using nth term formula, we get,

a20 = a+(20−1)d

a20 = 253+(19)(−5)

a20 = 253−95

a = 158

Therefore, 20th term from the last term of the AP 3, 8, 13, …, 253 is 158

Explanation:

nth term of the AP is;

an = a+(n−1)d

a4 = a+(4−1)d

a4 = a+3d

In the similar way, here we can present,

a8 = a+7d

a6 = a+5d

a10 = a+9d

Presented that,

a4+a8 = 24

a+3d+a+7d = 24

2a+10d = 24

a+5d = 12 …………………………………………………… (i)

a6+a10 = 44

a +5d+a+9d = 44

2a+14d = 44

a+7d = 22 …………………………………….. (ii)

On removing equation (i) from (ii),here  we achieve,

2d = 22 − 12

2d = 10

d = 5

By equation (i), we achieve,

a+5d = 12

a+5(5) = 12

a+25 = 12

a = −13

a2 = a+d = − 13+5 = −8

a3 = a2+d = − 8+5 = −3

Thus the first three terms of this A.P. are −13, −8, and −3.

Explanation:

According to the presented question, the incomes of Subba Rao exceedsevery year by Rs.200 and thus, forms an AP.

Later 1995, the salaries of each year are;

5000, 5200, 5400, …

First term, a = 5000

Common difference, d = 200

Later the  nth year, his salary will be Rs 7000.

nth term formula of AP,

an = a+(n−1) d

7000 = 5000+(n−1)200

200(n−1)= 2000

(n−1) = 10

n = 11

Thus in the 11th year, his salary is  Rs 7000.

Explanation:

Ramkali started off by saving Rs. 5 the first week before increasing his weekly savings by Rs. 1.75.

First term, a = 5

Common difference, d = 1.75

Presented,

an = 20.75

Find, n = ?

we notice, by the nth term formula,

an = a+(n−1)d

20.75 = 5+(n -1)×1.75

15.75 = (n -1)×1.75

(n -1) = 15.75/1.75 = 1575/175

= 63/7 = 9

n -1 = 9

n = 10

Explanation:

(i) In the Given, 2, 7, 12 ,…, to 10 terms

In A.P.,

The first term, a = 2

The  common difference, d = a2 − a1 = 7−2 = 5

n = 10

The formula for sum of nth term in AP series is,

Sn = n/2 [2a +(n-1)d]

S10 = 10/2 [2(2)+(10 -1)×5]

= 5[4+(9)×(5)]

= 5 × 49 = 245

(ii) In Given, −37, −33, −29 ,…, to 12 terms

In A.P.,

The first term, a = −37

The common difference, d = a2− a1

d= (−33)−(−37)

= − 33 + 37 = 4

n = 12

The formula for sum of nth term in AP series is,

Sn = n/2 [2a+(n-1)d]

S12 = 12/2 [2(-37)+(12-1)×4]

= 6[-74+11×4]

= 6[-74+44]

= 6(-30) = -180

(iii) In Given, 0.6, 1.7, 2.8 ,…, to 100 terms

In A.P.,

The first term, a = 0.6

The Common difference, d = a2 − a1 = 1.7 − 0.6 = 1.1

n = 100

The formula for sum of nth term in AP series is,

Sn = n/2[2a +(n-1)d]

S12 = 50/2 [1.2+(99)×1.1]

= 50[1.2+108.9]

= 50[110.1]

= 5505

(iv) In Given, 1/15, 1/12, 1/10, …… , to 11 terms

In A.P.,

The First term, a = 1/5

The Common difference, d = a2 –a1 = (1/12)-(1/5) = 1/60

And the number of terms n = 11

The formula for sum of nth term in AP series is,

Sn = n/2 [2a + (n – 1) d]

= 11/2(2/15 + 10/60)

= 11/2 (9/30)

= 33/20

(i)

The first terma=7

nth term, an = 84

Let 84 be the nth term of this A.P., 

an = a(n-1)d

84 = 7+(n – 1)×7/2

77 = (n-1)×7/2

22 = n−1

n = 23

The sum of n term is;

Sn = n/2 (a + l) , l = 84

Sn = 23/2 (7+84)

Sn  = (23×91/2) = 2093/2

(ii) 34 + 32 + 30 + ……….. + 10

In  this A.P.,

The first term, a = 34

The common difference, d = a2−a1 = 32−34 = −2

nth term, an= 10

Let 10 be the nth term in this A.P., ,

an = a +(n−1)d

10 = 34+(n−1)(−2)

−24 = (n −1)(−2)

12 = n −1

n = 13

The  sum of n terms is;

Sn = n/2 (a +l) , l = 10

= 13/2 (34 + 10)

= (13×44/2) = 13 × 22

= 286

(iii) (−5) + (−8) + (−11) + ………… + (−230)

In this A.P.,

The First term, a = −5

nth term, an= −230

The Common difference, d = a2−a1 = (−8)−(−5)

⇒d = − 8+5 = −3

Let −230 be the nth term of this A.P., ,

an= a+(n−1)d

−230 = − 5+(n−1)(−3)

−225 = (n−1)(−3)

(n−1) = 75

n = 76

The Sum of n term,

Sn = n/2 (a + l)

= 76/2 [(-5) + (-230)]

= 38(-235)

= -8930

Explanation:

(i) In Given , a = 5, d = 3, an = 50

From the formula of the nth term in an AP,

an = a +(n −1)d,

Here we putting the given values, we get,

⇒ 50 = 5+(n -1)×3

⇒ 3(n -1) = 45

⇒ n -1 = 15

⇒ n = 16

The sum of n terms,

Sn = n/2 (a +an)

Sn = 16/2 (5 + 50) = 440

(ii) In Given , a = 7, a13 = 35

From the formula of the nth term in an AP,

an = a+(n−1)d,

Here we  putting the given values, we get,

⇒ 35 = 7+(13-1)d

⇒ 12d = 28

⇒ d = 28/12 = 2.33

Now, Sn = n/2 (a+an)

S13 = 13/2 (7+35) = 273

(iii) In Given , a12 = 37, d = 3

From the formula of the nth term in an AP,

an = a+(n −1)d,

Here we  putting the given values, we get,

⇒ a12 = a+(12−1)3

⇒ 37 = a+33

⇒ a = 4

The sum of nth term,

Sn = n/2 (a+an)

Sn = 12/2 (4+37)

= 246

(iv) In Given , a3 = 15, S10 = 125

From the formula of the nth term in an AP,

an = a +(n−1)d,

Here  putting the given values, we get,

a3 = a+(3−1)d

15 = a+2d ………………………….. (i)

The Sum of the nth term,

Sn = n/2 [2a+(n-1)d]

S10 = 10/2 [2a+(10-1)d]

125 = 5(2a+9d)

25 = 2a+9d ……………………….. (ii)

Multiplying equation (i) by (ii), we will get;

30 = 2a+4d ………………………………. (iii)

By subtracting equation (iii) from (ii), we get,

−5 = 5d

d = −1

From the  equation (i),

15 = a+2(−1)

15 = a−2

a = 17 

First term

a10 = a+(10−1)d

a10 = 17+(9)(−1)

a10 = 17−9 = 8

(v) In Given that, d = 5, S9 = 75

The sum of n terms in AP is,

Sn = n/2 [2a +(n -1)d]

The sum of first nine terms are;

S9 = 9/2 [2a +(9-1)5]

25 = 3(a+20)

25 = 3a+60

3a = 25−60

a = -35/3

Here  the nth term can be written as;

an = a+(n−1)d

a9 = a+(9−1)(5)

= -35/3+8(5)

= -35/3+40

= (35+120/3) = 85/3

(vi) In Given , a = 2, d = 8, Sn = 90

The  sum of n terms in an AP is,

Sn = n/2 [2a +(n -1)d]

90 = n/2 [2a +(n -1)d]

⇒ 180 = n(4+8n -8) = n(8n-4) = 8n2-4n

⇒ 8n2-4n –180 = 0

⇒ 2n2–n-45 = 0

⇒ 2n2-10n+9n-45 = 0

⇒ 2n(n -5)+9(n -5) = 0

⇒ (n-5)(2n+9) = 0

So, n = 5 (as n only be a positive integer)

∴ a5 = 8+5×4 = 34

(vii) In Given , a = 8, an = 62, Sn = 210

The  sum of n terms in an AP is,

Sn = n/2 (a + an)

210 = n/2 (8 +62)

⇒ 35n = 210

⇒ n = 210/35 = 6

Now, 62 = 8+5d

⇒ 5d = 62-8 = 54

⇒ d = 54/5 = 10.8

(viii) In Given , nth term, an = 4, common difference, d = 2, sum of n terms, Sn = −14.

From the formula of the nth term in an AP,

an = a+(n −1)d,

Here we, putting the given values, we get,

4 = a+(n −1)2

4 = a+2n−2

a+2n = 6

a = 6 − 2n …………………………………………. (i)

The sum of n terms is;

Sn = n/2 (a+an)

-14 = n/2 (a+4)

−28 = n (a+4)

−28 = n (6 −2n +4) {From equation (i)}

−28 = n (− 2n +10)

−28 = − 2n2+10n

2n2 −10n − 28 = 0

n2 −5n −14 = 0

n2 −7n+2n −14 = 0

n (n−7)+2(n −7) = 0

(n −7)(n +2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

However, n can neither be negative nor fractional.

Hence, n = 7

From equation (i), we get

a = 6−2n

a = 6−2(7)

= 6−14

= −8

(ix) In Given the first term, a = 3,

Number of terms, n = 8

The  sum of n terms, S = 192

And  we know,

Sn = n/2 [2a+(n -1)d]

192 = 8/2 [2×3+(8 -1)d]

192 = 4[6 +7d]

48 = 6+7d

42 = 7d

d = 6

(x) In Given , l = 28,S = 144 and there are total of 9 terms.

The Sum of n terms formula,

Sn = n/2 (a + l)

144 = 9/2(a+28)

(16)×(2) = a+28

32 = a+28

a = 4

Explanation

Given  n terms of the AP. 9, 17, 25 …

In  this A.P.,

The First term, a = 9

The Common difference, d = a2−a1 = 17−9 = 8

The sum of n terms, is;

Sn = n/2 [2a+(n -1)d]

636 = n/2 [2×a+(8-1)×8]

636 = n/2 [18+(n-1)×8]

636 = n [9 +4n −4]

636 = n (4n +5)

4n2 +5n −636 = 0

4n2 +53n −48n −636 = 0

n (4n + 53)−12 (4n + 53) = 0

(4n +53)(n −12) = 0

Either 4n+53 = 0 or n−12 = 0

n = (-53/4) or n = 12

n cannot be negative or fraction, hence, n = 12 only

Explanation:

In Given 

The first term, a = 5

andlast term, l = 45

Sum of the AP, Sn = 400

And we know, the sum of AP formula is;

Sn = n/2 (a+l)

400 = n/2(5+45)

400 = n/2(50)

Number of terms, n =16

And  we know, the last term of AP series can be written as;

l = a+(n −1)d

45 = 5 +(16 −1)d

40 = 15d

The Common difference, d = 40/15 = 8/3

Explanation:

In Given ,

The First term, a = 17

And Last term, l = 350

The Common difference, d = 9

Here  n terms in the A.P., the formula for last term is

l = a+(n −1)d

350 = 17+(n −1)9

333 = (n−1)9

(n−1) = 37

n = 38

Sn = n/2 (a+l)

S38 = 38/2 (17+350)

= 19×367

= 6973

Therefore  this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

Explanation:

In Given,

The Common difference, d = 7

22nd term, a22 = 149

The Sum of first 22 term, S22 = ?

The formula of nth term,

an = a+(n−1)d

a22 = a+(22−1)d

149 = a+21×7

149 = a+147

a = 2 = First term

The Sum of n terms,

Sn = n/2(a+an)

S22 = 22/2 (2+149)

= 11×151

= 1661

Explanation:

In Given 

As Second term, a2 = 14

Third term, a3 = 18

And Common difference, d = a3−a2 = 18−14 = 4

a2 = a+d

14 = a+4

a = 10 = First term

The  Sum of n terms;

Sn = n/2 [2a + (n – 1)d]

S51 = 51/2 [2×10 (51-1) 4]

= 51/2 [20+(50)×4]

= 51 × 220/2

= 51 × 110

= 5610

Explanation:

In Given ,

S7 = 49

S17 = 289

The  Sum of n terms;

Sn = n/2 [2a + (n – 1)d]

Hence,

S7= 7/2 [2a +(n -1)d]

S7 = 7/2 [2a + (7 -1)d]

49 = 7/2 [2a +6d]

7 = (a+3d)

a + 3d = 7 …………………………………. (i)

In the same way,we get

S17 = 17/2 [2a+(17-1)d]

289 = 17/2 (2a +16d)

17 = (a+8d)

a +8d = 17 ………………………………. (ii)

Subtracting equation (i) from equation (ii),

5d = 10

d = 2

From equation (i), we can write it as a;

a+3(2) = 7

a+ 6 = 7

a = 1

Therefore

Sn = n/2[2a+(n-1)d]

= n/2[2(1)+(n – 1)×2]

= n/2(2+2n-2)

= n/2(2n)

= n2

Explanation:

(i) Given that

 an = 3+4n

a1 = 3+4(1) = 7

a2 = 3+4(2) = 3+8 = 11

a3 = 3+4(3) = 3+12 = 15

a4 = 3+4(4) = 3+16 = 19

The common difference between the terms are;

a2 − a1 = 11−7 = 4

a3 − a2 = 15−11 = 4

a4 − a3 = 19−15 = 4

Therefore , ak + 1 − ak is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.

The sum of nth term is;

Sn = n/2[2a+(n -1)d]

S15 = 15/2[2(7)+(15-1)×4]

= 15/2[(14)+56]

= 15/2(70)

= 15×35

= 525

(ii) an = 9−5n

a1 = 9−5×1 = 9−5 = 4

a2 = 9−5×2 = 9−10 = −1

a3 = 9−5×3 = 9−15 = −6

a4 = 9−5×4 = 9−20 = −11

The common difference between the terms are;

a2 − a1 = −1−4 = −5

a3 − a2 = −6−(−1) = −5

a4 − a3 = −11−(−6) = −5

Therefore, ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

The sum of nth term is;

Sn = n/2 [2a +(n-1)d]

S15 = 15/2[2(4) +(15 -1)(-5)]

= 15/2[8 +14(-5)]

= 15/2(8-70)

= 15/2(-62)

= 15(-31)

= -465

Explanation:

In Given ,

Sn = 4n−n2

The First term, a = S1 = 4(1) − (1)2 = 4−1 = 3

The Sum of first two terms = S2= 4(2)−(2)2 = 8−4 = 4

And Second term, a2 = S2 − S1 = 4−3 = 1

The Common difference, d = a2−a = 1−3 = −2

Nth term, an = a+(n−1)d 

= 3+(n −1)(−2)

= 3−2n +2

= 5−2n

Hence, a3 = 5−2(3) = 5-6 = −1

a10 = 5−2(10) = 5−20 = −15

Hence, the sum of the first two terms is 4. The second phrase is 1.

The third, tenth, and nth terms, respectively, are 1, 15, and 5 2n.

Explanation:

The positive integers 6, 12, 18, and 24 are all divisible by 6.

First word in this series of forms of an A.P. is six, and common difference is six. a = six.

d = 6

S40 = ?

 Formula of sum of n terms, ,

Sn = n/2 [2a +(n – 1)d]

Here  putting n = 40, we will  get,

S40 = 40/2 [2(6)+(40-1)6]

= 20[12+(39)(6)]

= 20(12+234)

= 20×246

= 4920

Explanation:

There are eight multiples: 8, 16, 24, 32.

This series follows the AP format, with the first term being 8 and the common difference being 8.

Hence, a = 8

d = 8

S15 = ?

The formula of sum of nth term, we know that

Sn = n/2 [2a+(n-1)d]

S15 = 15/2 [2(8) + (15-1)8]

= 15/2[16 +(14)(8)]

= 15/2[16 +112]

= 15(128)/2

= 15 × 64

= 960.

Explanation:

Between 0 to 50, the odd numbers are 1, 3, 5, 7, 9,... 49.

We are aware that the form of odd numbers is A.P.

Therefore,

The First term, a = 1

And Common difference, d = 2

The Last term, l = 49

Formula of last term is,

l = a+(n−1) d

49 = 1+(n−1)2

48 = 2(n − 1)

n − 1 = 24

n = 25 = Number of terms

The formula  sum of nth term,

Sn = n/2(a +l)

S25 = 25/2 (1+49)

= 25(50)/2

=(25)(25)

= 625.

Explanation:

The fines are shown as an A.P. with a first term of 200 and a common difference of 50.

A = 200 and D = 50 as a result.

The fine must be paid. if the contractor delayed the project by 30 days, the answer is S30.

The sum of the nth term formula

Sn = n/2[2a+(n -1)d]

Hence,

S30= 30/2[2(200)+(30 – 1)50]

= 15[400+1450]

= 15(1850)

= 27750

Here the contractor has to pay Rs 27750 as penalty.

Explanation:

The first award is worth Rs. P.

Prize of second place = Rs. P 20

The price of third place is Rs. P  − 40.

These awards have an A.P. cost, with the first word being P and the common difference being 20.

Thus, d = -20 and a = P

Provided, S7 equals 700.

We are aware of the sum of the nth term formula:

Sn = n/2 [2a + (n – 1)d]

7/2 [2a + (7 – 1)d] = 700

a + 3(−20) = 100

a −60 = 100

a = 160

As a result, the rewards were worth Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs. 60, and Rs. 40, respectively.

Explanation:

We can count the amount of trees that AP students have planted.

1, 2, 3, 4, 5………………..12

The First term, a = 1

And Common difference, d = 2−1 = 1

Sn = n/2 [2a +(n-1)d]

S12 = 12/2 [2(1)+(12-1)(1)]

= 6(2+11)

= 6(13)

= 78

78 trees were planted by one section of the classes.

3 sections of the classes planted a total of 234 trees, or 3 x 78.

Thus, the kids will plant 234 trees.

Explanation:

Thus, we know that the circumference of a semicircle equals r.

P1 = π(0.5) = π/2 cm

P2 = π(1) = π cm

P3 = π(1.5) = 3π/2 cm

 P1, P2, P3 are the lengths of the semi-circles.

Here  we got a series  as,

π/2, π, 3π/2, 2π, ….

P1 = π/2 cm

P2 = π cm

The Common difference, d = P2 – P1 = π – π/2 = π/2

The First term = P1= a = π/2 cm

The formula for the sum of n terms is Sn = n/2 [2a + (n - 1)d].

The total circumference of 13 successive circles is;

S13 = 13/2 [2(π/2) + (13 – 1)π/2]

=  13/2 [π + 6π]

=13/2 (7π)

= 13/2 × 7 × 22/7

= 143 cm

Explanation:

The log counts in each row are expressed as an A.P.20, 19, 18...

First term in the supplied A.P. is 20 and common difference is d = a2-a1 = 19-20 = -1.

200 logs in total will be arranged in n rows.

 Sn = 200

Total of nth term formula,

Sn = n/2 [2a +(n -1)d]

S12 = 12/2 [2(20)+(n -1)(-1)]

400 = n (40−n+1)

400 = n (41-n)

400 = 41n−n2

n2−41n + 400 = 0

n2−16n−25n+400 = 0

n(n −16)−25(n −16) = 0

(n −16)(n −25) = 0

(n −16) = 0 or n−25 = 0

n = 16 or n = 25

 nth term formula is ,

an = a+(n−1)d

a16 = 20+(16−1)(−1)

a16 = 20−15

a16 = 5

The 25th term  written as;

a25 = 20+(25−1)(−1)

a25 = 20−24

= −4

As the numbers cannot be negative, there are 5 logs on the 16th row.

Hence, 200 logs can be arranged in 16 rows, with 5 logs in the 16th row.

Explanation:

In terms of AP, the potato distances from the bucket are 5, 8, 11, 14, etc.

The competitor's distance to gather the potatoes is double the distance at which they have previously been maintained.

the distances to be run in relation to the distances of the potatoes as

10, 16, 22, 28, 34,……….

Thus, the first term, a = 10 and d = 16−10 = 6

S10 =?

Formula of sum of n terms is,,

S10 = 10/2 [2(10)+(10 -1)(6)]

= 5[20+54]

= 5(74)

= 370

The competitor will run 370 metres in total.

Explanation:

AP series is 121, 117, 113, . . .,

The, first term, a = 121

AndCommon difference, d = 117-121= -4

The nth term formula,

an = a+(n −1)d

Hence,

an = 121+(n−1)(-4)

= 121-4n+4

=125-4n

We have to find the first negative term of the series, an < 0

Thus,

125-4n < 0

125 < 4n

n>125/4

n>31.25

Hence, the 32nd term in the series is the first negative term.

Explanation:

In  the given statements, we have to write

a3 + a7 = 6 …………………………….(i)

And

a3 ×a7 = 8 ……………………………..(ii)

 nth term of the formula,

an = a+(n−1)d

Third term, a3 = a+(3 -1)d

a3 = a + 2d………………………………(iii)

And Seventh term, a7= a+(7-1)d

a7 = a + 6d ………………………………..(iv)

by equation (iii) and (iv),we  putting in equation(i), we will get

a+2d +a+6d = 6

2a+8d = 6

a+4d=3

or

a = 3–4d …………………………………(v)

reputting the eq.(iii) and (iv), in eq. (ii), we will get

(a+2d)×(a+6d) = 8

Putting the value of a from equation (v), ,

(3–4d +2d)×(3–4d+6d) = 8

(3 –2d)×(3+2d) = 8

32 – 2d2 = 8

9 – 4d2 = 8

4d2 = 1

d = 1/2 or -1/2

We  putting both the values of d, 

a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1, when d = 1/2

a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5, when d = -1/2

The sum of nth term of AP is;

Sn = n/2 [2a +(n – 1)d]

So, when a = 1 and d=1/2

The sum of first 16 terms are;

S16 = 16/2 [2 +(16-1)1/2] = 8(2+15/2) = 76

And when a = 5 and d= -1/2

The sum of first 16 terms are;

S16 = 16/2 [2(5)+(16-1)(-1/2)] = 8(5/2)=20

Explanation:

The ladder has rungs that are 25 cm apart.

And the ladder's top to lowest rung distance is equal to

= 5/2 ×100cm

= 250cm

The  total number of rungs = 250/25 + 1 = 11

The illustration demonstrates how the ladder's rungs are positioned from top to bottom in decreasing height. We can now infer that the rungs are arranged in ascending AP order.

The length of wood required for the rungs will be calculated based on the sum of the phrases in an AP series.

So,

The First term, a = 45

And Last term, l = 25

The Number of terms, n = 11

 the  sum of nth terms is equal to,

Sn= n/2(a+ l)

Sn= 11/2(45+25) = 11/2(70) = 385 cm

Hence  the length of the wood required for the rungs is 385 cm.

Explanation:

Row house numbers range from 1, 2, 3, 4, 5, and 49.

The row of houses with numbers on them is an AP shape.

So,

The First term, a = 1

And Common difference, d=1

We know  the number of xth houses can be written  as;

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

Sum of number of houses beyond x house = Sx-1

= (x-1)/2[2(1)+(x-1-1)1]

= (x-1)/2 [2+x-2]

= x(x-1)/2 ………………………………………(i)

By the given condition, we get

S49 – Sx = {49/2[2(1)+(49-1)1]}–{x/2[2(1)+(x-1)1]}

= 25(49) – x(x + 1)/2 ………………………………….(ii)

As per the  condition, eq.(i) and eq(ii) are equal to each other;

Hence

x(x-1)/2 = 25(49) – x(x+1)/2

x = ±35

 The number of houses cannot be a negative number. 

Therefore, the value of x is 35.

Explanation:

The first step in the supplied figure is 1/2 m wide, the second step is 1 m wide, and the third step is 3/2 m wide. Here, it is clear that each step's width is 1/2 m when its height is 1/4 m. Moreover, the steps are always 50m long. Hence, an AP series is formed by the step width.

½ , 1, 3/2, 2, ……..Volume of steps = Volume of Cuboid

= Length × Breadth X Height

Now,

The first phase requires a volume of concrete equal to  = ¼ ×1/2 ×50 = 25/4

Therefore the amount of concrete needed to construct the second step is ¼ ×1×50 = 25/2

The second step's concrete volume requirement is ¼ ×3/2 ×50 = 75/4

Now that we know how much concrete is needed to make the steps, we can see that it is 25/4, 25/2, and 75/4 in the AP series.

Here , applying the AP series concept,

The First term, a = 25/4

And Common difference, d = 25/2 – 25/4 = 25/4

The sum of n terms is;

Sn = n/2[2a+(n-1)d] = 15/2(2×(25/4 )+(15/2 -1)25/4)

solving, we get,

Sn = 15/2 (100)

Sn=750

The terrace requires 750 m3 of concrete in its entirety.