Question

-7 Formula of sum of n terms of an APj​

Answer 1

Answer:

Sn=n/2[2a+(n-1)d]

Explanation:

Answer 2

Explanation :

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

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

• 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 of an A.P.,

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

where

    a - first term

    d - common difference

    n - number of terms

     l - last term

   aₙ - nth term

   Sₙ - sum of n terms

Derivation :

a₁ = a

a₂ = a + d

a₃ = a₂ + d = a + d + d = a + 2d

....

aₙ = a + (n - 1)d

⇒ Sum of first n terms,

Sₙ = a₁ + a₂ + a₃ + ..... + aₙ

Sₙ = a + (a + d) + (a + 2d) + ..... + (a + (n-1)d) ---[1]

________________________

aₙ = aₙ

aₙ₋₁ = aₙ - d

aₙ₋₂ = aₙ₋₁ - d = aₙ - d - d = aₙ - 2d

.....

a₁ = aₙ + (n - 1)(-d) = aₙ - (n - 1)d

⇒ Sum of n terms from the end,

Sₙ = aₙ + aₙ₋₁ + aₙ₋₂ + ..... + a₁

Sₙ = aₙ + (aₙ - d) + (aₙ - 2d) + ...... aₙ - (n - 1)d --- [2]

Add both the equations,

Sₙ + Sₙ = a + (a + d) + (a + 2d) +.... + [a +(n-1)d] + aₙ + (aₙ - d) + (aₙ - 2d) + .... + (aₙ - (n-1)d)

2Sₙ = (a + aₙ) + (a + d + aₙ - d) + (a + 2d + aₙ - 2d) + ..... + (a + (n-1)d + aₙ - (n-1)d)

2Sₙ = (a + aₙ) + (a + aₙ) + (a + aₙ) + .... + (a + aₙ)

2Sₙ = n(a + aₙ)

 Sₙ = [n(a + aₙ)]/2

We know,

nth term , aₙ = a + (n - 1)d

 Sₙ = (n/2) [a + a + (n - 1)d]

 Sₙ = (n/2) [2a + (n - 1)d]