NCERT Grade 10-Arithmetic Progressions-Answers

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1.    (i) Taxi fare for 1st km = Rs 15, Taxi fare after 2 km = 15 + 8 = Rs 23

Taxi fare after 3 km = 23 + 8 = Rs 31

Taxi fare after 4 km = 31 + 8 = Rs 39

Therefore, the sequence is 15, 23, 31, 39…

It is an arithmetic progression because difference between any two consecutive terms is equal which is 8. (23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, …)

(ii) Let amount of air initially present in a cylinder = 

Amount of air left after pumping out air by vacuum pump

Amount of air left when vacuum pump again pumps out air

So, the sequence we get is like 

Checking for difference between consecutive terms …

Difference between consecutive terms is not equal.

Therefore, it is not an arithmetic progression.

(iii) Cost of digging 1 meter of well = Rs 150

Cost of digging 2 meters of well = 150 + 50 = Rs 200

Cost of digging 3 meters of well = 200 + 50 = Rs 250

Therefore, we get a sequence of the form 150, 200, 250…

It is an arithmetic progression because difference between any two

consecutive terms is equal. (200 – 150 = 250 – 200 = 50….)

Here, difference between any two consecutive terms which is also called common difference is equal to 50.

(iv) Amount in bank after 1st year =      …. (1)

Amount in bank after two year =         ….. (2)

Amount in bank after three year =       …..(3)

Amount in bank after four year =       ….(4)

It is not an arithmetic progression because (2) – (1) ≠ (3) – (2)

(Difference between consecutive terms is not equal)

Therefore, it is not an Arithmetic Progression.

2.     (i) First term = a = 10, d = 10

Second term = a + d = 10 + 10 = 20

Third term = second term + d = 20 + 10 = 30

Fourth term = third term + d = 30 + 10 = 40

Therefore, first four terms are: 10, 20, 30, 40

(ii) First term = a = -2 , d = 0

Second term = a + d = -2 + 0 = -2

Third term = second term + d = -2 + 0 = -2

Fourth term = third term + d = -2 + 0 = -2

Therefore, first four terms are: -2, -2, -2, -2

(iii) First term = a = 4, d =-3

Second term = a + d = 4 – 3 = 1

Third term = second term + d = 1 – 3 = -2

Fourth term = third term + d = -2 – 3 = -5

Therefore, first four terms are: 4, 1, -2, -5

(iv) First term = a = -1, d = 1/2

Second term = a + d = -1 + 1/2 = -1/2

Third term = second term + d = –1/2 + 1/2 = 0

Fourth term = third term + d = 0 + 1/2 = 1/2

Therefore, first four terms are: -1, -1/2, 0, 1/2

(v) First term = a = -1.25, d = -0.25

Second term = a + d = -1.25 – 0.25 = -1.50

Third term = second term + d = -1.50 – 0.25 = -1.75

Fourth term = third term + d = –1.75 – 0.25 = -2.00

Therefore, first four terms are: -1.25, -1.50, -1.75, -2.00

3.    (i) 3, 1, -1, -3…

First term = a = 3,

Common difference (d) = Second term – first term = Third term – second term

and so on

Therefore, Common difference (d) = 1- 3 = -2

(ii) -5, -1, 3, 7… First term = a = -5

Common difference (d) = Second term – First term

= Third term – Second term and so on

Therefore, Common difference (d) = -1 – (-5) = -1 + 5 = 4

(iii) 

First term = a = 

Common difference (d) = Second term – First term

= Third term – Second term and so on

Therefore, Common difference (d)

(iv) 0.6, 1.7, 2.8, 3.9…

First term = a = 0.6

Common difference (d) = Second term – First term

= Third term – Second term and so on

Therefore, Common difference (d) = 1.7 – 0.6 = 1.1

4.    (i) 2, 4, 8, 16…

It is not an AP because difference between consecutive terms is not equal.

As   4 – 2 ≠ 8 – 4

(ii) 

It is an AP because difference between consecutive terms is equal.

Common difference (d) = 1/2

Fifth term =                          Sixth term = 4 + 1/2 = 

Seventh term = 

Therefore, next three terms are 4,  and 5.

(iii) -1.2, -3.2, -5.2, -7.2…

It is an AP because difference between consecutive terms is equal.

⇒ -3.2 – (-1.2) = -5.2 – (-3.2) = -7.2 – (-5.2) = -2

Common difference (d) = -2

Fifth term = -7.2 – 2 = -9.2                  Sixth term = -9.2 – 2 = -11.2

Seventh term = -11.2 – 2 = -13.2

Therefore, next three terms are -9.2, -11.2 and -13.2

(iv) -10, -6, -2, 2…

It is an AP because difference between consecutive terms is equal.

⇒ -6 – (-10) = -2 – (-6) = 2 – (-2) = 4

Common difference (d) = 4

Fifth term = 2 + 4 = 6                            Sixth term = 6 + 4 = 10

Seventh term = 10 + 4 = 14

Therefore, next three terms are 6, 10 and 14

(v) 3, 3 + √2, 3 + 2 √2, 3 + 3 √2……

It is an AP because difference between consecutive terms is equal.

⇒ 3 + √2 – 3 = √2, 3 + 2√2 – (3+√2) = 3 + 2√2 – 3 – √2 =√2

Common difference (d) = -√2

Fifth term = 3 + 3√2 + √2 = 3 + 4√2

Sixth term = 3 + 4√2 + -√2 = 3 + 5√2

Seventh term = 3 + 5√2 + √2 = 3 + 6√2

Therefore, next three terms are (3 + 4 √2), (3 + 5√2), (3 + 6√2)

(vi) 2, 0.22, 0.222, 0.2222…

It is not an AP because difference between consecutive terms is not equal.

⇒ 0.22 – 0.2 ≠ 0.222 – 0.22

(vii) 0, -4, -8, -12…

It is an AP because difference between consecutive terms is equal.

-4 – 0 = -8 – (-4) = -12 – (-8) = -4

Common difference (d) = -4

Fifth term = -12 – 4 =-16                     Sixth term = -16 – 4 = -20

Seventh term = -20 – 4 = -24

Therefore, next three terms are -16, -20 and -24

(viii) 

It is an AP because difference between consecutive terms is equal.

(ix) 1, 3, 9, 27…

It is not an AP because difference between consecutive terms is not equal.

⇒  3 – 1 ≠ 9 – 3

(x) a, 2a, 3a, 4a…

It is an AP because difference between consecutive terms is equal.

⇒ 2a – a = 3a – 2a = 4a – 3a = a

Common difference (d) = a

Fifth term = 4a + a = 5a                       Sixth term = 5a + a = 6a

Seventh term = 6a + a = 7a

Therefore, next three terms are 5a, 6a and 7a

(xi) a, a2, a3, a4

It is not an AP because difference between consecutive terms is not equal. a2 – a a3 – a2

⇒ a– a ≠ a3 – a2

(xii) √2, √8, √18, √32…..     ⇒      √2, 2√2, 3√2, 4√2

It is an AP because difference between consecutive terms is equal.

⇒ 2√2 – √2 = 3√2 – 2√2 = √2

Common difference (d) = √2

Fifth term = 4√2 + √2 = 5√2             Sixth term = 5√2 + √2 = 6√2

Seventh term = 6√2 + √2 = 7√2

Therefore, next three terms are 5√2, 6√2, 7√2

(xiii) √3, √6, √9, √12…..

It is not an AP because difference between consecutive terms is not equal.

⇒ √6 – √3 ≠ √9 – √6

(xiv) 12, 32, 52, 72

It is not an AP because difference between consecutive terms is not equal.

⇒32 – 12 ≠ 52 – 32

(xv) 12, 52, 72, 73…          ⇒          1, 25, 49, 73…

It is an AP because difference between consecutive terms is equal.

⇒ 52 – 12 ≠ 72 – 52 ≠ 73 – 72 ≠ 24

Common difference (d) = 24

Fifth term = 73 + 24 = 97                     Sixth term = 97 + 24 = 121

Seventh term = 121 + 24 = 145

Therefore, next three terms are 97, 121 and 145

5.    (i) a = 7, d = 3, n = 8

We need to find a, here.

Using formula an = a + (n – 1) d

Putting values of a, d and n,

an= 7 + (8 – 1) 3 = 7 + (7) 3 = 7 + 21 = 28

(ii) a = -18, n= 10, a= 0

We need to find d here.

Using formula an = a + (n – 1) d

Putting values of a, an and n,

     0 = -18 + (10 – 1) d

⇒ 0 = -18 + 9d               ⇒         18 = 9d          ⇒            d = 2

(iii) d = -3, n = 18, an = -5

We need to find a here.

Using formula an = a + (n – 1) d

Putting values of d, an and n,

    -5 = a + (18 – 1) (-3)

⇒ -5 = a + (17) (-3)   ⇒    -5=a-51     ⇒     a = 46

(iv) a = -18.9, d = 2.5, an = 3.6 We need to find n here.

Using formula an = a + (n – 1) d

Putting values of d, an and a,

    3.6 = -18.9 + (n – 1) (2.5)

⇒ 3.6 = -18.9 + 2.5n – 2.5            ⇒            2.5n = 25         ⇒           n = 10

(v) a = 3.5, d = 0, n = 105

We need to find a,, here.

Using formula an = a + (n – 1) d

Putting values of d, n and a,

    an = 3.5 + (105 – 1) (0)

⇒ an = 3.5 + 104 x 0           ⇒             an = 3.5 + 0         ⇒              an = 3.5

6.    (i) 10, 7, 4…

First term = a = 10, Common difference = d = 7 – 10 = 4 – 7 = -3

And n = 30 {Because, we need to find 30th term}

     an = a + (n – 1) d

⇒ a30 = 10 + (30 – 1) (-3) = 10 – 87 = -77

Therefore, the answer is (C).

(ii) -3, -1/2, 2…

First term = a = -3, Common difference = d = 

And n = 11 (Because, we need to find 11th term)

Therefore 11th term is 22 which means answer is (B).

7.    (i) 2, __, 26

We know that difference between consecutive terms is equal in any A.P. Let the missing term be x.

x – 2 = 26 – x            ⇒          2x = 28             ⇒            x = 14

Therefore, missing term is 14.

(ii) __, 13, __, 3

Let missing terms be x and y.

The sequence becomes x, 13, y, 3

We know that difference between consecutive terms is constant in any A.P.

y – 13 = 3 -y            ⇒                 2y = 16          ⇒              y = 8

And 13 – x = y- 13            ⇒            x + y = 26

But, we have y = 8,

⇒ x + 8 = 26                   ⇒                       x = 18

Therefore, missing terms are 18 and 8.

(iii) 5, __, __, 

Here, first term = a = 5            And, 4th term = a4 =

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

Therefore, we get common difference = d = 

Second term = a + d =

Third term = second term + d =

Therefore, missing terms are  and 8.

(iv) -4, __, __, __, __, 6

Here, First term = a = -4        and      6th term = a6 = 6

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

a6 = -4 + (6 – 1) d       ⇒        6 = -4 + 5d

⇒ 5d = 10                 ⇒         d = 2

Therefore, common difference = d = 2

Second term = first term + d = a + d = -4+ 2 = -2

Third term = second term + d = -2 + 2 = 0

Fourth term = third term + d = 0 + 2 = 2

Fifth term = fourth term + d = 2 + 2 = 4

Therefore, missing terms are -2, 0, 2 and 4.

(v) __, 38, __, __, __, -22

We are given 2nd and 6th term.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

a2 = a + (2 – 1) d           And     a6 = a + (6 – 1) d

⇒ 38 = a + d                 And     -22 = a + 5d

These are equations in two variables, we can solve them using any method.

Using equation (38 = a + d), we can say that a = 38 – d.

Putting value of a in equation (-22 = a + 5d),

       -22 = 38 – d + Sd          ⇒            4d = -60                      d = -15

Using this value of d and putting this in equation 38 = a + d,

         38 = a – 15                               a = 53

Therefore, we get a = 53 and d = -15

First term = a = 53

Third term = second term + d = 38 – 15 = 23

Fourth term = third term + d = 23 – 15 = 8

Fifth term = fourth term + d = 8 – 15 = -7

Therefore, missing terms are 53, 23, 8 and -7.

8.    First term = a = 3, Common difference = d = 8 – 3 = 13 – 8 = 5 and an = 78

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

a, = 3 + (n – 1) 5,                     78 = 3 + (n – 1) 5                       75 = 5n – 5

⇒ 80 = 5n                                    n = 16

It means 16th term of the given AP is equal to 78.

9.    (i) 7, 13, 19 …, 205

First term = a = 7, Common difference = d = 13 – 7 = 19 – 13 = 6

And an = 205

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

205 = 7 + (n – 1) 6 = 7 + 6n – 6                         205 = 6n + 1

⇒ 204 = 6n      ⇒         n = 34

Therefore, there are 34 terms in the given arithmetic progression.

(ii) 18,  13….., -47

First term = a =18, Common difference = d = 

And an = -47

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

⇒ -94 = 36 – 5n + 5                5n = 135                n = 27

Therefore, there are 27 terms in the given arithmetic progression.

10.  Let -150 is the nth of AP 11, 8, 5, 2… which means that an = -150 Here, First term = a = 11, Common difference = d = 8 – 11 = -3

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

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

⇒ 3n = 164                        ⇒          

But, n cannot be in fraction.

Therefore, our supposition is wrong. -150 cannot be term in AP.

11.   Here a11= 38 and a16 = 73

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

38 = a + (11 – 1) (d)         And                 73 = a + (16 – 1) (d)

⇒ 38 = a + 10d                And                  73 = a + 15d

These are equations consisting of two variables.

We have, 38 = a + 10d                          a = 38 – 10d

Let us put value of a in equation (73 = a + 15d),

73 = 38 – 10d + 15d                             35 = 5d

Therefore, Common difference = d = 7

Putting value of d in equation 38 = a + 10d,

38 = a + 70                            a = -32

Therefore, common difference = d = 7 and First term = a = -32

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

a31 = -32 + (31 – 1) (7) = -32 + 210 = 178

Therefore, 31st term of AP is 178.

12.  An AP consists of 50 terms and the 50th term is equal to 106 and a3 = 12 Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

a50 = a + (50 – 1) d                 And                 a3 = a + (3 – 1) d

106 = a + 49d                       And                12 = a + 2d

These are equations consisting of two variables.

Using equation 106 = a + 49d, we get a = 106 – 49d

Putting value of a in the equation 12 = a + 2d,

12 = 106 – 49d + 2d                   47d = 94                   d = 2

Putting value of d in the equation, a = 106 – 49d,

a = 106 – 49 (2) = 106 – 98 = 8

Therefore, First term = a = 8 and Common difference = d = 2

To find 29th term, we use formula an = a + (n – 1) d which is used to find nth term of arithmetic progression,

a29 = 8 + (29 – 1) 2 = 8 + 56 = 64

Therefore, 29th term of AP is equal to 64.

13.   It is given that 3rd and 9th term of AP are 4 and -8 respectively. It means a3 = 4 and a9 = -8

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

4 = a + (3 – 1) d          And,       -8 = a + (9 – 1) d

⇒ 4 = a + 2d              And,       -8 = a + 8d

These are equations in two variables.

Using equation 4 = a + 2d, we can say that a = 4 – 2d

Putting value of a in other equation -8 = a + 8d,

-8 = 4 – 2d + 8d

⇒ -12 = 6d                       d = -2

Putting value of d in equation -8 = a + 8d,

-8 = a + 8 (-2)       -8=a-16           a = 8
Therefore, first term = a = 8 and Common Difference = d = -2 We want to know which term is equal to zero.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

0 = 8 + (n – 1) (-2)           0 = 8 – 2n + 2                     0 = 10 – 2n

⇒ 2n = 10                                    n = 5

Therefore, 5th term is equal to 0.

14.  a11 = a10 + 7    … (1)

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

a17 = a + 16d                     … (2)

a10 = a + 9d                               (3)

Putting (2) and (3) in equation (1),

a + 16d = a + 9d + 7              7d = 7     ⇒       d = 1

15.  Lets first calculate 54th of the given AP.

First term = a = 3, Common difference = d = 15 – 3 = 12

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

a54 = a + (54 – 1) d = 3 + 53 (12) = 3 + 636 = 639

We want to find which term is 132 more than its 54th term.

Let us suppose it is nth term which is 132 more than 54th term.

an = a54 + 132                 ⇒          3 + (n 1) 12 = 639 + 132

⇒ 3 + 12n – 12 = 771                    12n 9 = 771

⇒ 12n = 780                                  n = 65

Therefore, 65th term is 132 more than its 54th term.

16.   Let first term of 1st AP = a

Let first term of 2nd AP = a’

It is given that their common difference is same.

Let their common difference be d.

It is given that difference between their 100th terms is 100.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

a + (100 – 1) d – [a’ + (100 – 1) d] = a + 99d – a’ – 99d = 100

⇒ a – a’ = 100                  … (1)

We want to find difference between their 1000th terms which means we want to calculate:

a + (1000 – 1) d – [a’ + (1000 – 1) d] = a + 999d – a’ – 999d = a – a’

Putting equation (1) in the above equation,

a + (1000 – 1) d – [a’ + (1000 – 1) d] = a + 999d – a’ + 999d = a – a’ = 100

Therefore, difference between their 1000th terms would be equal to 100.

17.   We have AP starting from 105 because it is the first three digit number divisible by 7. AP will end at 994 because it is the last three digit number divisible by 7. Therefore, we have AP of the form 105, 112, 119…, 994

Let 994 is the nth term of AP.

We need to find n here.

First term = a = 105, Common difference = d = 112 – 105 = 7

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

994 = 105 + (n – 1) (7)                          994 = 105 + 7n – 7

⇒ 896 = 7n                              ⇒               n = 128

It means 994 is the 128th term of AP.

Therefore, there are 128 terms in AR

18.  First multiple of 4 which lie between 10 and 250 is 12.

The last multiple of 4 which lie between 10 and 250 is 248.

Therefore, AP is of the form 12, 16, 20… ,248

First term = a = 12, Common difference = d = 4

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

248 = 12 + (n – 1) (4)                           248 = 12 + 4n 4

⇒ 240 = 4n                                              n = 60

It means that 248 is the 60th term of AP.

So, we can say that there are 60 multiples of 4 which lie between 10 and 250.

19.   Lets first consider AP 63, 65, 67…

First term = a = 63, Common difference = d = 65 – 63 = 2

Using formula an = a + (n – d, to find nth term of arithmetic progression,

a, = 63 + (n – 1) (2)                 … (1)

Now, consider second AP 3, 10, 17…

First term = a = 3, Common difference = d = 10 – 3 = 7

Using formula an = a + (n – d, to find nth term of arithmetic progression,

an = 3 + (n – 1) (7)                              … (2)

According to the given condition:

(1) = (2)

⇒ 63 + (n – 1) (2) = 3 +(n1) (7)                   63 + 2n – 2 = 3 + 7n – 7

⇒ 65 = 5n                                                         n = 13

Therefore, 13th terms of both the AP’s are equal.

20.  Let first term of AP = a

Let common difference of AP = d

It is given that its 3rd term is equal to 16.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

16 = a + (3 – 1) (d)       ⇒       16 = a + 2d                      …(1)

It is also given that 7th term exceeds 5th term by 12.

According to the given condition:

a7 = as + 12                 a + (7 – 1) d = a + (5 – 1) d + 12

⇒ 2d = 12                       d = 6

Putting value of d in equation 16 = a + 2d,

16 = a + 2 (6)                a = 4

Therefore first term = a = 4

And, common difference = d = 6

Therefore, AP is 4, 10, 16, 22…

21.   We want to find 20th term from the last term of given AP.

So, let us write given AP in this way: 253 … 13, 8, 3

Here First term = a = 253, Common Difference = d = 8 – 13 = -5

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

a20 = 253 + (20 – 1) (-5)      ⇒       a20 = 253 + 19 (-5) = 253 – 95 = 158

Therefore, the 20th term from the last term of given AP is 158.

22.  The sum of 4th and 8th terms of an AP is 24 and sum of 6th and 10th terms is 44. a4 + a= 24 and a6 + a10 = 44

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

a + (4 – 1) d + [a + (8 – 1) d] = 24

And, a + (6 – 1) d + [a + (10 – 1) d] = 44

⇒ a + 3d + a + 7d = 24 And a + 5d + a + 9d = 44

⇒ 2a + 10d = 24                      And 2a + 14d = 44

⇒ a + 5d = 12                          And a + 7d =22

These are equations in two variables.

Using equation a + Sd = 12, we can say that a = 12 – 5d               …(1)

Putting (1) in equation a + 7d = 22,

12 – 5d + 7d = 22                                  12 + 2d = 22

⇒ 2d = 10                                               d = 5

Putting value of d in equation a = 12 – 5d,

a = 12 – 5 (5) = 12 – 25 = -13

Therefore, first term = a = -13 and, Common difference = d = 5

Therefore, AP is -13, -8, -3, 2…

Its first three terms are -13, -8 and -3.

23.  Subba Rao’s starting salary = Rs 5000

It means, first term = a = 5000

He gets an increment of Rs 200 after every year.

Therefore, common difference = d = 200

His salary after 1 year = 5000 + 200 = Rs 5200

His salary after two years = 5200 + 200 = Rs 5400

Therefore, it is an AP of the form 5000, 5200, 5400, 5600… , 7000 We want to know in which year his income reaches Rs 7000.

Using formula an= a + (n – 1) d, to find nth term of arithmetic progression,

7000 = 5000 + (n – 1) (200)                        7000 = 5000 + 200n – 200

⇒ 7000 – 5000 + 200 = 200n        ⇒              2200 = 200n

⇒ n = 11

It means after 11 years, Subba Rao’s income would be Rs 7000.

24.   Ramkali saved Rs. 5 in the first week of year. It means first term = a = 5

Ramkali increased her weekly savings by Rs 1.75.

Therefore, common difference = d = Rs 1.75

Money saved by Ramkali in the second week = a + d = 5 + 1.75 = Rs 6.75

Money saved by Ramkali in the third week = 6.75 + 1.75 = Rs 8.5

Therefore, it is an AP of the form: 5, 6.75, 8.5 , 20.75

We want to know in which year her weekly savings become 20.75.

Using formula an = a + (n – d, to find nth term of arithmetic progression,

      20.75 = 5 + (n – 1) (1.75)                      20.75 = 5 + 1.75n – 1.75

⇒ 17.5 = 1.75n                                                 n = 10

It means in the 10th week her savings become Rs 20.75.

25.  (i) 2, 7, 12… to 10 terms

Here First term = a = 2, Common difference = d = 7 – 2 = 5 and n = 10

Applying formula,  to find sum of n terms of AP,

(ii) -37, -33, -29… to 12 terms

Here First term = a = -37, Common difference = d = -33 – (-37) = 4 And n = 12

Applying formula,  to find sum of n terms of AP,

(iii) 6, 1.7, 2.8… to 100 terms

Here First term = a = 0.6, Common difference = d = 1.7 – 0.6 = 1.1 And n = 100

Applying formula,  to find sum of n terms of AP,

(iv)  to 11 terms

Here First tern = a =    Common difference = d 

Applying formula,  to find sum of n terms of AP,

26.  (i) 

Here First term = a = 7, Common difference = 

And Last term = l = 84

We do not know how many terms are there in the given AP. So, we need to find n first.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

[7 + (n – 1) (3.5)] = 84                         7 + (3.5) n – 3.5 = 84

⇒ 3.5n = 84 + 3.5 – 7                    3.5n = 80.5

⇒ n = 23

Therefore, there are 23 terms in the given AR

It means n = 23.

Applying formula,  to find sum of n terms of AP,

(ii) 34 + 32 + 30 +….. + 10

Here First term = a = 34, Common difference = d = 32 – 34 = -2 And Last term = l = 10

We do not know how many terms are there in the given AP.

So, we need to find n first.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

      [34 + (n – 1) (-2)] = 10                    34 – 2n + 2 = 10

⇒ -2n = -26                                               n = 13

Therefore, there are 13 terms in the given AP.

It means n = 13.

Applying formula,  to find sum of n terms of AP,

(iii) -5 + (-8) + (-11) + + (-230)

Here First term = a = -5, Common difference = d = -8 – (-5) = -8 + 5 = -3 And Last term = l = -230

We do not know how many terms are there in the given AP.

So, we need to find n first.

Using formula an = a + (n -1) d, to find nth term of arithmetic progression,

[-5 + (n 1) (-3)] = -230         ⇒              -5-3n+3=-230

⇒  -3n = -228                          ⇒              n = 76

Therefore, there are 76 terms in the given AP.

It means n = 76.

Applying formula,  to find sum of n terms of AP,

27.  (i) Given a = 5, d = 3, an = 50, find n and Sn.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

      an=5+(n-1) (3)                    50 = 5 + 3n – 3

⇒   48 = 3n                      ⇒          n = 16

Applying formula,  to find sum of n terms of AP,

Therefore, n = 16 and Sn = 440

(ii) Given a = 7, a13 = 35, find d and S13.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

     a13 = 7 + (13 – (d)          ⇒             35 = 7 + 12d

⇒ 28 = 12d                                      

Applying formula,  to find sum of n terms of AP,

(iii) Given a12 = 37, d = 3, find a and S12.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression, a12 = a + (12 – 1) 3         37 = a + 33       ⇒       a = 4

Applying formula,  to find sum of n terms of AP,

Therefore a = 4 and S12 = 246

(iv) Given a3 = 15, S10 = 125, find d and a10.

Using formula an = a + (n – d, to find nth term of arithmetic progression,

a3 = a + (3 – 1) (d)                      ⇒                   15 = a + 2d

⇒ a = 15 – 2d    …(1)

Applying formula,  to find sum of n terms of AP,


Putting (1) in the above equation,

125 = 5 {2 (15 – 2d) + 9d] = 5 (30 – 4d + 9d)

⇒ 125 = 150 + 25d 125 – 150 = 25d

⇒ -25 = 25d

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression, a10 = a + (10 – 1) d

Putting value of d and equation (1) in the above equation,

a10⇒ = 15 – 2d + 9d = 15 + 7d = 15 + 7 (-1) = 15 – 7 = 8

Therefore, d = -1 and a10 = 8

(v) Given d = 5, S9 = 75, find a and a9.

Applying formula,  to find sum of n terms of AP,

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

(vi) Given a = 2, d = 8, Sn= 90, find n and an.

Applying formula,  to find sum of n terms of AP,

We discard negative value of n because here n cannot be in negative or fraction. The value of n must be a positive integer.

Therefore, n = 5

Using formula an = a + (n – d, to find nth term of arithmetic progression,

a5 = 2 + (5 – 1) (8) = 2 + 32 = 34

Therefore, n = 5 and an = 34

(vii) Given a = 8, an = 62, Sn = 210, find n and d.

Using formula an = a + (n -1) d, to find nth term of arithmetic progression,

62 = 8 + (n – 1) (d) = 8 + nd – d          ⇒         62 = 8 + nd – d

⇒ nd – d = 54                            ⇒                   nd = 54 + d     … (1)

Applying formula,  to find sum of n terms of AP,

Putting equation (1) in the above equation,

(viii) Given an = 4, d = 2, Sn = -14, find n and a.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

4 = a + (n – 1) (2) = a + 2n – 2

⇒ 4 = a + 2n – 2                                     6 = a + 2n

⇒  a = 6 – 2n         … (1)

Applying formula,  to find sum of n terms of AP,

Putting equation (1) in the above equation, we get

-28 = n [2 (6 – 2n) + 2n – 2]                  -28 = n (12 – 4n + 2n — 2)

⇒ -28 = n (10 – 2n)                                     2n2 – lOn – 28 = 0

⇒ n2 – 5n – 14 = 0                                            n2 – 7n + 2n – 14 = 0

⇒ n (n – 7) + 2 (n – 7) = 0                 ⇒        (n + 2) (n – 7) = 0

⇒ n = -2, 7

Here, we cannot have negative value of n.

Therefore, we discard negative value of n which means n = 7. Putting value of n in equation (1), we get

a = 6 – 2n = 6 – 2 (7) = 6 – 14 = -8

Therefore, n = 7 and a = -8

(ix) Given a = 3, n = 8, S = 192, find d.

Using formula,  to find sum of n terms of AP, we get

(x) Given I = 28, S = 144, and there are total of 9 terms. Find a.

Applying formula,  , to find sum of n terms, we get

28.  First term = a = 9, Common difference = d = 17 – 9 = 8, Sn = 636

Applying formula,  to find sum of n terms of AP, we get

Comparing equation 4n2 + 5n – 636 = 0 with general form an2 + bn + c = 0, we get a = 4, b = 5 and c = -636

We discard negative value of n here because n cannot be in negative, n can only be a positive integer.

Therefore, n = 12

Therefore, 12 terms of the given sequence make sum equal to 636.

29.  First term = a = 5, Last term = 1 = 45, Sn = 400

Applying formula,   to find sum of n terms of AP, we get

Applying formula,  to find sum of n terms of AP and putting value of n, we get

30.  First term = a = 17, Last term = 1 = 350 and Common difference = d = 9

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression, we get

350 = 17 + (n – 1) (9)               350 = 17 + 9n – 9

⇒ 342 = 9n                        ⇒         n = 38

Applying formula,  to find sum of n terms of AP and putting value of n, we get

Therefore, there are 38 terms and their sum is equal to 6973.

31.   It is given that 22nd term is equal to 149 ⇒  a22 = 149

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression, we get

149 = a + (22 – 1) (7)                 149 = a + 147          a = 2

Applying formula,  to find sum of n terms of AP and putting value of n, we get

Therefore, sum of first 22 terms of AP is equal to 1661.

32.  It is given that second and third term of AP are 14 and 18 respectively.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression, we get

14 = a + (2 – 1) d                   ⇒                14 = a + d        …(1)

And, 18 = a + (3 – 1) d        ⇒               18 = a + 2d … (2)

These are equations consisting of two variables.

Using equation (1), we get, a = 14 – d

Putting value of a in equation (2), we get

18 = 14 – d + 2d                 ⇒                     d=4

Therefore, common difference d= 4

Putting value of d in equation (1), we get

18 = a + 2 (4)              a = 10

Applying formula,  to find sum of n terms of AP, we get

Therefore, sum of first 51 terms of an AP is equal to 5610.

33.  It is given that sum of first 7 terms of an AP is equal to 49 and sum of first 17 terms is equal to 289.

Applying formula,  to find sum of n terms of AP, we get

Putting equation (1) in the above equation, we get

17 = 7 – 3d + 8d                       10 = 5d                   d=2

Putting value of d in equation (1), we get

a = 7 – 3d = 7 – 3 (2) = 7 – 6 = 1

Again apply formula,  to find sum of n terms of AP, we get

Therefore, sum of n terms of AP is equal to n2.

34.  (i) We need to show that a1, a2 … an form an AP where an = 3 + 4n

Let us calculate values of a1, a2, a3 … using an = 3 + 4n

a1 = 3 + 4 (1) = 3 + 4 = 7                      a2 = 3 + 4 (2) = 3 + 8 = 11

a3 = 3 + 4 (3) = 3 + 12 = 15                 a4 = 3 + 4 (4) = 3 + 16 = 19

So, the sequence is of the form 7, 11, 15, 19 …

Let us check difference between consecutive terms of this sequence.

11 – 7 = 4,     15 – 11 = 4,     19 – 15 = 4

Therefore, the difference between consecutive terms is constant which means terms a1, a2an form an AR

We have sequence 7, 11, 15, 19 …

First term = a = 7 and Common difference = d = 4

Applying formula,  to find sum of n terms of AP, we get

Therefore, sum of first 15 terms of AP is equal to 525.

(ii) We need to show that a1, a2an form an AP where an = 9 – 5n

Let us calculate values of a1, a2, a3 … using 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

So, the sequence is of the form 4, -1, -6, -11 …

Let us check difference between consecutive terms of this sequence.

-1 – (4) = -5,           -6 – (-1) = -6 + 1 = -5,           -11 – (-6) = -11 + 6 = -5

Therefore, the difference between consecutive terms is constant which means terms a1, a2an form an AP.

We have sequence 4, -1, -6, -11 …

First term = a = 4 and Common difference = d = -5

Applying formula,  to find sum of n terms of AP, we get

Therefore, sum of first 15 terms of AP is equal to -465.

35.  It is given that the sum of n terms of an AP is equal to (4n – n2)
It means Sn = 4n – n2

Let us calculate Si and S2 using Sn = 4n – n2

S1 = 4 (1) – (1)2 = 4 – 1 = 3

S2 = 4 (2) – (2)2 = 8 – 4 = 4

First term = a = S= 3              … (1)

Let us find common difference now.

We can write any AP in the form of general terms like a , a + d, a + 2d …

We have calculated that sum of first two terms is equal to 4 i.e. S2 = 4

Therefore, we can say that a + (a + d) = 4

Putting value of a from equation (1), we get

2a + d = 4                                 2 (3) + d = 4

⇒ 6 + d = 4                              d = -2

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression,

Second term of AP = a2 = a + (2 – 1) d = 3 + (2 – 1) (-2) = 3 – 2 = 1

Third term of AP = a3 = a + (3 – 1) d = 3 + (3 – 1) (-2) = 3 – 4 = -1

Tenth term of AP = a10 = a + (10 – 1) d = 3 + (10 – 1) (-2) = 3 – 18 = -15

nth term of AP = an = a + (n – 1) d = 3 + (n – 1) (-2) = 3 – 2n + 2 = 5 – 2n

36.  The first 40 positive integers divisible by 6 are 6, 12, 18, 24 … 40 terms.
Therefore, we want to find sum of 40 terms of sequence of the form:

6, 12, 18, 24 … 40 terms

Here, first term = a = 6 and Common difference = d = 12 – 6 = 6, n = 40

Applying formula,  to find sum of n terms of AP, we get

37.  The first 15 multiples of 8 are 8, 16, 24, 32 … 15 times

First term = a = 8 and Common difference = d = 16 – 8 = 8, n = 15

Applying formula,  to find sum of n terms of AP, we get

38.  The odd numbers between 0 and 50 are 1, 3, 5, 7 … 49

It is an arithmetic progression because the difference between consecutive terms is constant.

First term = a = 1, Common difference = 3 – 1 = 2, Last term = / = 49

We do not know how many odd numbers are present between 0 and 50. Therefore, we need to find n first.

Using formula an = a + (n – 1) d, to find nth term of arithmetic progression, we get

49 = 1 + (n – 1) 2                              49 = 1 + 2n – 2

⇒ 50 = 2n                         ⇒                 n = 25

Applying formula,  to find sum of n terms of AP, we get

39.  Penalty for first day = Rs 200, Penalty for second day = Rs 250

Penalty for third day = Rs 300

It is given that penalty for each succeeding day is Rs 50 more than the preceding day.

It makes it an arithmetic progression because the difference between consecutive terms is constant.

We want to know how much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

So, we have an AP of the form 200, 250, 300, 350 … 30 terms

First term = a = 200, Common difference = d = 50, n = 30

Applying formula,  to find sum of n terms of AP, we get

⇒   Sn = 15 (400 + 1450) = 27750

Therefore, penalty for 30 days is Rs. 27750.

40.  It is given that sum of seven cash prizes is equal to Rs 700.

And, each prize is R.s 20 less than its preceding term.

Let value of first prize = Rs. a

Let value of second prize =Rs (a – 20)

Let value of third prize = Rs (a – 40)

So, we have sequence of the form:

a, a – 20, a – 40, a – 60 …

It is an arithmetic progression because the difference between consecutive terms is constant.

First term = a, Common difference = d = (a – 20) – a = -20

n = 7 (Because there are total of seven prizes)

S7 = Rs 700 {given}

Applying formula,  to find sum of n terms of AP, we get

Therefore, value of first prize = Rs 160

Value of second prize = 160 – 20 = Rs 140

Value of third prize = 140 – 20 = Rs 120

Value of fourth prize = 120 – 20 = Rs 100

Value of fifth prize = 100 – 20 = Rs 80

Value of sixth prize = 80 – 20 = Rs 60

Value of seventh prize = 60 – 20 = Rs 40

41.  There are three sections of each class and it is given that the number of trees planted

by any class is equal to class number.

The number of trees planted by class I = number of sections × 1 = 3 × 1 = 3 The number of trees planted by class II = number of sections × 2 = 3 × 2 = 6 The number of trees planted by class III = number of sections × 3 = 3 × 3 = 9 Therefore, we have sequence of the form 3, 6, 9 … 12 terms

To find total number of trees planted by all the students, we need to find sum of the sequence 3, 6, 9, 12 … 12 terms.

First term = a = 3, Common difference = d= 6 – 3 = 3 and n = 12

Applying formula,  to find sum of n terms of AP, we get

42.  

Length of semi circle of radii 0.5 cm = π (0.5) cm

Length of semi circle of radii 1.0 cm = π (1.0) cm

Length of semi circle of radii 1.5 cm = π (1.5) cm

Therefore, we have sequence of the form:

π (0.5), π (1.0), π (1.5) … 13 terms {There are total of thirteen semi-circles}.

To find total length of the spiral, we need to find sum of the sequence π (0.5), π (1.0), π (1.5) … 13 terms

Total length of spiral = π (0.5) + π(1.0) + π(1.5) … 13 terms

Total length of spiral = π (0.5 + 1.0 + 1.5) … 13 terms        … (1)

Sequence 0.5, 1.0, 1.5 … 13 terms is an arithmetic progression.

Let us find the sum of this sequence.

First term = a = 0.5, Common difference = 1.0 – 0.5 = 0.5 and n = 13

Applying formula,  to find sum of n terms of AP, we get

Therefore, 0.5 + 1.0 + 1.5 + 2.0 … 13 terms = 45.5

Putting this in equation (1), we get

Total length of spiral = π (0.5 + 1.5 + 2.0 + … 13 terms) = π (45.5) = 143 cm

43. The number of logs in the bottom row = 20

The number of logs in the next row = 19

The number of logs in the next to next row = 18

Therefore, we have sequence of the form 20, 19, 18 …

First term = a = 20, Common difference = d = 19 – 20 = -1

We need to find that how many rows make total of 200 logs.

Applying formula,  to find sum of n terms of AP, we get

It is a quadratic equation, we can factorize to solve the equation.

⇒ n2– 25n – 16n + 400 = 0                        n (n – 25) – 16 (n – 25) = 0

⇒ (n – 25) (n – 16)                           ⇒                  n = 25, 16

We discard n = 25 because we cannot have more than 20 rows in the sequence. The sequence is of the form: 20, 19, 18 …

At most, we can have 20 or less number of rows.

Therefore, n = 16 which means 16 rows make total number of logs equal to 200. We also need to find number of logs in the 16th row.

Applying formula,  to find sum of n terms of AP, we get

200 = 8 (20 + l)                   ⇒                200 = 160 + 8l

⇒ 40 = 8l                              ⇒                  l = 5

Therefore, there are 5 logs in the top most row and there are total of 16 rows.

44.  The distance of first potato from the starting point = 5 meters

Therefore, the distance covered by competitor to pickup first potato and put it in bucket = 5 × 2 = 10 meters

The distance of Second potato from the starting point = 5 + 3 = 8 meters {All the potatoes are 3 meters apart from each other}

Therefore, the distance covered by competitor to pickup 2nd potato and put it in bucket = 8 × 2 = 16 meters

The distance of third potato from the starting point = 8 + 3 = 11 meters

Therefore, the distance covered by competitor to pickup 3rd potato and put it in bucket = 11 × 2 = 22 meters

Therefore, we have a sequence of the form 10, 16, 22 … 10 terms (There are ten terms because there are ten potatoes)

To calculate the total distance covered by the competitor, we need to find : 10 + 16 + 22 + … 10 terms

First term = a = 10, Common difference = d = 16 – 10 = 6

n = 10 {There are total of 10 terms in the sequence}

Applying formula,  to find sum of n terms of AP, we get

Therefore, total distance covered by competitor is equal to 370 meters.

45.  Given: 121, 117, 113, ..

Here        a =121, d =117-121= 4

Now,        an a+(n-1)d

= 121+(n-1)(-4) = 121-4n+4 = 125-4n

For the first negative term, an < 0

Hence, the first negative term is 32nd term.

46.  Let the AP be a-4d, a-3d, a-2d, a-d, a, a+d, a+2d, a+3d,   …. Then, a3 = a -2d, a7 = a +2d

⇒  a3+ a= a – 2d + a + 2d = 6

⇒  2a = 6                    a = 3 ………………….. (i)

Also                 (a -2d)(a +2d)= 8

⇒  a2 – 4d2 = 8                        4d2 = a2 -8

47.   Number of rungs

The length of the wood required for rungs = sum of 10 rungs

48.   Here a =1 and d =1

 

According to question,

Since, x is a counting number, so negative value will be neglected.

x =35

49.  Volume of concrete required to build the first step, second step, third step,     (in m2)

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