Question-Mathematics - Arithmetic Progressions - Ask NCERT class 0 with solution questions answer. All ncert questions Answers for class 0 subject Mathematics chapter and topic Arithmetic Progressions

**Question 1.** **Asked on :**30 October 2022:03:14:04 AM

If m times of m^{th} term of an A.P. is equal to n times of n^{th} term of same A.P. Show that (m + n)^{th} term is zero.

*-Added by ATP Admin* See More Answers **Mathematics** » **Arithmetic Progressions**

**Answer:**

Let the first term of A.P. is a and common difference d = d,

अत: m(a_{m}) = n(a_{n})

**⇒** m[a + (m – 1)d] = n[a + (n – 1)d]

**⇒** am + m(m – 1)d = an + n(n – 1)d

**⇒** am – an + [m^{2} – m – n^{2 }+ n]d = 0

**⇒** a(m – n ) + [m^{2} – n^{2 }–( m – n)]d = 0

**⇒** a(m – n ) + [(m + n) (m – n) – ( m – n)]d = 0

**⇒** (m – n ) [a + {(m + n)– 1}]d = 0

**⇒** a + {(m + n)– 1}d = 0 ………….. (1)

a_{m + n }= a + {(m + n)– 1}d

a_{m + n }= 0 From Equation (i)

*-Answered by ATP Admin* On 30 October 2022:03:17:09 AM

**Question 2.** **Asked on :**09 June 2022:11:09:09 AM

If the 1st term of a series is 7 and 13th term is 35. find the sum of 13 terms of the sequence.

*-Added by Master Purushottam* See More Answers **Mathematics** » **Arithmetic Progressions **

**Answer:**

a= 7 , a_{13} = 35 _{ }

S_{n} = n2 (a + l)

S_{13} = 132 (7+35)

= 132 x 42

= 13 x 21

= 273

Hence, the sum of the first 13 term of the sequence is 273.

*-Answered by Master Purushottam* On 09 June 2022:11:19:25 AM

**Question 3.** **Asked on :**01 June 2022:05:35:09 PM

In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?

*-Added by Shruti Srivastva* See More Answers **Mathematics** » **Arithmetic Progressions **

**Answer:**

A.P. 23,21,19.....5

a=23, d= -2

a_{n} =5

=>a+(n-1)d=5

=>23+(n-1)-2=5

=>(n-1)-2=5-23

=>(n-1)=-18/-2

=>n=9

*-Answered by Shruti Srivastva* On 01 June 2022:06:01:47 PM

**Question 4.** **Asked on :**01 June 2022:05:25:29 PM

If s_{n}, the sum of first n terms of an A.P. is given by s_{n}=3n^{2}-4n, find the nth term.

*-Added by Shruti Srivastva* See More Answers **Mathematics** » **Arithmetic Progressions **

**Answer:**

s_{1}=a_{1}=3(1)^{2}-4(1)

=> 3-4

=> -1

s_{2}=3(2)^{2}-4(2)

=>12-8

=>4

s_{2}-s_{1}=a_{2}

a_{2}=4-(-1)

a_{2}=5

d=a_{2}-a_{1}

d=5-(-1)

d=6

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

=> -1+(n-1)6

=> -1+6n-6

=>6n-7

*-Answered by Shruti Srivastva* On 01 June 2022:05:28:25 PM

**Question 5.** **Asked on :**01 June 2022:05:16:00 PM

Which term of the A.P. 3,15,27,39..... will be 120 more than its 21st term?

*-Added by Shruti Srivastva* See More Answers **Mathematics** » **Arithmetic Progressions **

**Answer:**

a=3, d=12, n=21

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

=> 3+(21-1)12

=> 3+(20)12

=> 3+240

=> 243

120 more than 21st term =243+120=363

a_{n} = 363

=>a+(n-1)d=363

=>3+(n-1)12=363

=>(n-1)12=363-3

=>(n-1)=360/12

=>n=30+1

=> n=31

*-Answered by Shruti Srivastva* On 01 June 2022:05:19:13 PM

**Question 6.** **Asked on :**01 June 2022:04:22:08 PM

The sum of 5th term and 9th term of an A.P is 30. If its 25th term is three times its 8th term, find the A.P.

*-Added by Shruti Srivastva* See More Answers **Mathematics** » **Arithmetic Progressions **

**Answer:**

Given that:

a_{5} + a_{9} = 30

⇒ a + 4d + a + 8d = 30

⇒ 2a + 12d = 30

⇒ 2(a + 6d) = 30

⇒ a + 6d = 15 ............ (i)

a_{25} = 3(a_{8}) (given)

⇒ a + 24d = 3(a + 7d)

⇒ a + 24d = 3a + 21d

⇒ 3a - a + 21d - 24d = 0

⇒ 2a - 3d = 0

⇒ 2a = 3d

a = 3d2 ............ (ii)

Putting the value of a in equation (i)

⇒ a + 6d = 15

⇒ 3d2 + 6d = 15

= 3d + 12d2 = 15

15d = 30

d = 3015

d = 2

Putting the value d = 2 in equation (ii)

a = 3 x 22 = 3

a = 3

The required A.P. Will be

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

3, 3 + 2, 3 + 2x2, 3 + 3x2, ....................

3, 5, 7, 9 .................... Answer

*-Answered by Master Mind* On 27 October 2022:05:20:23 PM

**Question 7.** **Asked on :**01 June 2022:04:12:59 PM

Find the number of natural numbers between101 and 999 which are divisible bu both 2 and 5.

*-Added by Shruti Srivastva* See More Answers **Mathematics** » **Arithmetic Progressions **

**Answer:**

Number which are divisible by 2 and 5

=> number which are diisible by 2×5=10

A.P. 110,120,130.....990

a=110, d=10

a_{n} = 990

=>a+(n-1)d=990

=> 110+(n-1)10=990

=> (n-1)10=990-110

=> (n-1)10=880

=> (n-1)=880/10

=> n=88+1

=> n=89

*-Answered by Shruti Srivastva* On 01 June 2022:04:16:31 PM

**Question 8.** **Asked on :**01 June 2022:03:56:13 PM

The sum of 5th and 9th terms of an A.P. is 72 and the sum of 7th and 12th term is 97. Find the A.P.

*-Added by Shruti Srivastva* See More Answers **Mathematics** » **Arithmetic Progressions **

**Answer:**

a_{5}+a_{9} =72

=> a+4d+a+8d=72

=> 2a+12d=72

=> a+6d=36

=> a=36-6d----(1)

a_{7}+a_{12} = 97

=> a+6d+a+11d=97

=> 2a+17d=97----(2)

putting eq(1) in eq(2)

=>2a+17d=97

=>2(36-6d)+17d=97

=>72-12d+17d=97

=>5d=97-72

=>5d=25

=>d=25/5

=>d=5

putting d in eq.(1)

a=36-6d

a=36-6(5)

a=36-30

a=6

A.P. 6,11,16,21....

*-Answered by Shruti Srivastva* On 01 June 2022:03:58:33 PM

**Question 9.** **Asked on :**01 June 2022:03:44:51 PM

Find the sum of integers between 10 and 500 which are divisible by7.

*-Added by Shruti Srivastva* See More Answers **Mathematics** » **Arithmetic Progressions **

**Answer:**

A.P., 14,21............497

a=14, d=7

a_{n} = 497

=>a+(n-1)d=497

=>14+(n-1)7=497

=>(n-1)7=497-14

=>(n-1)=483/7

=>(n-1)=69

=>n=69+1

=>n=70

s_{n} = n/2[2a+(n-1)d]

=> 70/2[2×14+(70-1)7]

=> 35[28+483]

=> 35[511]

=> 17885

*-Answered by Shruti Srivastva* On 01 June 2022:03:46:36 PM

**Question 10.** **Asked on :**01 June 2022:03:33:05 PM

Find the middle terms of the A.P. 7,13,19...241.

*-Added by Shruti Srivastva* See More Answers **Mathematics** » **Arithmetic Progressions **

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