Question

find the sum of first 50 terms of the series 1,3,5,7.....​

Answer 1

Answer:

The sum of first 50 terms of the AP is 2500

Step-by-step explanation:

Given terms of an AP are: 1, 3, 5, 7...

The first term i.e, a is = 1

Common Difference  = 7 - 5 = 2

To find the sum of given number of terms in an AP, we use the formula:-

$S{}_{n}=\dfrac{n}{2}[2a+(n-1)d]$

Here, 'n' is the number of terms in the AP.

'a' represents the first term of the AP.

'd' represent the common difference.

$S{}_{n}$ represent the sum of 'n' number of terms of an AP.

Substitute the values in the formula:-

$S{}_{50}=\dfrac{50}{2}[2(1)+(50-1)2]$

$S{}_{50}=25[2+(49)2]$

$S{}_{50}=25[2+98]$

$S{}_{50}=25[100]$

$S{}_{50}=2500$

So, the sum of first 50 terms of the AP is 2500 !

$\large\boxed{\large\boxed{\large\boxed{Solved !}}}}$

Answer 2

$\huge\mathfrak\pink{HELLO}$

${Here's}$${your}$ ${answer}$

METHOD 1 :

First term = a₁ = 1

Second term = a₂ = 3

Common difference = d = a₂- a₁ = 3 - 1 = 2

n/2 = 50/2 = 25

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

Sn = 25 [ 2×1 + (50-1 ) 2 ]

=25 [2 + (49) 2 ]

=25 [2 + 98 ]

=25 [100]

=2500

The sum of 1st 50 terms of the sequence 1,3,5,7... is 2500

METHOD 2 :

Sum of first n odd Integers = n²

1 , 3 , 5 ........50 terms = ( 50 )²

= 2500