Mathematics - Arithmetic Progressions - Ask NCERT

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

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 + [m² – m – n² + n]d = 0

⇒ a(m – n ) + [m² – n² –( 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)

 a= 7 , a₁₃ = 35 
   

S_n = n2 (a + l)

S₁₃ = 132 (7+35)

      = 132 x 42

      = 13 x 21 

      = 273

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

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

s₁=a₁=3(1)²-4(1)

=> 3-4

=> -1

s₂=3(2)²-4(2)

=>12-8

=>4

s₂-s₁=a₂ 

a₂=4-(-1)

a₂=5

d=a₂-a₁

d=5-(-1)

d=6

a_n = a+(n-1)d

=> -1+(n-1)6

=> -1+6n-6

=>6n-7

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

Given that: 

a₅ + a₉ = 30 

⇒ a + 4d + a + 8d = 30

⇒ 2a + 12d = 30 

⇒ 2(a + 6d) = 30 

⇒ a + 6d = 15 ............ (i) 

a₂₅ = 3(a₈)  (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

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

a₅+a₉ =72

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

=> 2a+12d=72

=> a+6d=36

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

a₇+a₁₂ = 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....

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