Question

arithmetic sequence 3,5,7,9 . find the sum of the first n term of sequence​

Answer 1

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Given Arithmetic progression:

• 3, 5, 7, 9......

Let's identify its basic elements:

• First term (a) = 3
• Common difference (d) = 5 - 3 = 7 - 5 = 2

We have to find the sum of n terms of the arithmetic progression for which we can use the given formula:

${ \cdot{ \boxed{ \rm{sn = \dfrac{n}{2} \bigg( 2a + (n - 1)d \bigg)}}}}$

Let's plug the given values of a and d,

⇛ $\rm{sn = \dfrac{n}{2} \bigg(2 \times 3 + (n - 1)2 \bigg)}$

Simplifying,

⇛ $\rm{sn = \dfrac{n}{2} \bigg(6 + 2n - 2 \bigg)}$

⇛ $\rm{sn = \dfrac{n}{2} \bigg(2n + 4 \bigg)}$

This can be written as,

⇛ $\rm{sn = n(n + 2)}$

⇛ $\rm{sn = {n}^{2} + 2n}$

Thus, Sum of n terms of the AP is n² + 2n

Answer 2

Answer:

Given : -

• AP is 3 ,5 ,7 ....

a = 3

we have to find D

D = A_2 - A_1

Substitute all values :

D = 5 - 3 = 2

Now Apply Sn formula

→ Sn = n/2 ( 2a + (n-1 ) d )

Substitute all value :

→ sn = n/2 (2×3 + (n - 1) 2)

→ Sn = n/2 ( 6 + 2n - 2 )

→ Sn = n/2 (4 + 2n)

→ Sn = ( 4n + 2n² ) /2

→ Sn = 2n( 2 + n ) / 2

→ Sn = n( 2 + n )

→ Sn = n² + 2n