Question

for an AP. Sn =n²+1, find a1, a2, a3 and d​

Answer 1

Answer :

• a₁ = 1
• a₂ = 3
• a₃ = 5
• d = 2

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 , ..........

• nth term of AP,

          $\boxed{\bf a_n=a+(n-1)d}$

• Sum of n terms,

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

___________________________

Sum of n terms = n² + 1

Let's find the sum of (n - 1) terms.

Put n = n - 1,

Sum of (n - 1) terms = (n - 1)² + 1

                                = n² + 1² - 2(n)(1) + 1

                                = n² + 1 - 2n + 1

                                = n² - 2n + 2

nth term =  Sum of n terms - (sum of (n - 1) terms)

               =  n² + 1 - ( n² - 2n + 2 )

               =  n² + 1 - n² + 2n - 2

               =  2n - 1

Therefore,

nth term , aₙ = 2n - 1

First term,

Put n = 1,

a₁ = 2(1) - 1

    = 2 - 1

    = 1

Second term,

Put n = 2,

a₂ = 2(2) - 1

     = 4 - 1

     = 3

Third term,

Put n = 3,

a₃ = 2(3) - 1

    = 6 - 1

    = 5

Common difference,

d = a₂ - a₁

  = 3 - 1

  = 2

Answer 2

Answer:

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