Question

find the A.p whose nth term is tn= 2n^2-3n+1​

Answer 1

Answer:

helo guys its your answer

Step-by-step explanation:

Second - method:- Sn= sigma(2n+1) Sn= 2. sigma n +sigma 1 Sn= 2×n.(n+1)/2 +n ... nth term of an AP is generally given by Tn=a+(n-1) d. ... How do you find the sum of the first 20 terms

Answer 2

Answer :

0 , 3 , 10 , 21 ,....

Step-by-step explanation :

$\underline{\underline{\bf Arithmetic \ Progression:}}$

• It is the sequence of numbers such that the difference between any two successive numbers is constant.

• In AP,

      a - first term

      d - common difference

      n - number of terms

       l - last term

     aₙ - nth term

     Sₙ - sum of n terms

• General form of AP,

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

• Formulae :-

       nth term of AP,

          $\boxed{\bf a_n=a+(n-1)d}$

       Sum of n terms in AP,

         $\boxed{\bf S_n=\frac{n}{2}[2a+(n-1)d]} \\\\ \boxed{\bf S_n=\frac{n}{2}[a+l]}$

______________________________

Given,

nth term , tₙ = 2n² - 3n + 1

First term :

 Substitute n = 1,

t₁ = 2(1)² - 3(1) + 1

  = 2(1) - 3 + 1

  = 2 - 3 + 1

  = 3 - 3

  = 0

Second term :

Substitute n = 2,

t₂ = 2(2)² - 3(2) + 1

   = 2(4) - 6 + 1

   = 8 - 6 + 1

   = 2 + 1

   = 3

Third term :

Substitute n = 3,

t₃ = 2(3)² - 3(3) + 1

   = 2(9) - 9 + 1

   = 18 - 9 + 1

   = 9 + 1

   = 10

Fourth term :

Substitute n = 4,

t₄ = 2(4)² - 3(4) + 1

    = 2(16) - 12 + 1

    = 32 - 12 + 1

    = 20 + 1

    = 21

The A.P. is 0 , 3 , 10 , 21 , ....