The nth term of an ap …

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On putting n = 1, 2,3 ,…. in eq (1), For n = 1 (an) = - 4n + 15 1 = - 4(1) + 15   a1 = 11   For n = 2 a2 = - 4(2) + 15   a2 = - 8 + 15   a2 = 7   Common Difference , d = a2 - a1   d = 7-11   d = - 4   By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d] S20 = (20/2)[2 × 11 + (20 -1)(-4)] S20 = 10 [22 + 19 × - 4[ S20 = 10 [22 - 76] S20 = 10 × - 54 S20 = - 540   Hence,  the sum of first 20 terms of AP is  - 540.

nth term of an A.P, (a_n) = - 4n + 15……..(1)

On putting n = 1, 2,3 ,…. in eq (1),

For n = 1

(a_n) = - 4n + 15

1 = - 4(1) + 15  

a₁ = 11

For n = 2

a₂ = - 4(2) + 15  

a₂ = - 8 + 15  

a₂ = 7

Common Difference , d = a₂ - a₁  

d = 7-11  

d = - 4

By using the formula ,Sum of nth terms , S_n = n/2 [2a + (n – 1) d]

S₂₀ = (20/2)[2 × 11 + (20 -1)(-4)]

S₂₀ = 10 [22 + 19 × - 4[

S₂₀ = 10 [22 - 76]

S₂₀ = 10 × - 54

S₂₀ = - 540  

Hence,  the sum of first 20 terms of AP is  - 540.

 

Kis book ka h

Given :  

nth term of an A.P, (an) = - 4n + 15……..(1)

 

On putting n = 1, 2,3 ,…. in eq (1),

For n = 1

(an) = - 4n + 15

1 = - 4(1) + 15  

a1 = 11

 

For n = 2

a2 = - 4(2) + 15  

a2 = - 8 + 15  

a2 = 7

 

Common Difference , d = a2 - a1  

d = 7-11  

d = - 4

 

By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]

S20 = (20/2)[2 × 11 + (20 -1)(-4)]

S20 = 10 [22 + 19 × - 4[

S20 = 10 [22 - 76]

S20 = 10 × - 54

S20 = - 540  

Hence,  the sum of first 20 terms of AP is  - 540.