Question

the sum of first 20 terms of AP 1, 3, 5, 7, 9...is

Answer 1

The sum of first 20 terms of AP 1 , 3 , 5 , 7 , 9 . . . is 400

Given :

The AP 1 , 3 , 5 , 7 , 9 . . .

To find :

The sum of first 20 terms of AP 1 , 3 , 5 , 7 , 9 . . .

Formula :

Sum of first n terms of an arithmetic progression

$\displaystyle \sf = \frac{n}{2} \bigg[2a + (n - 1)d \bigg]$

Where First term = a

Common Difference = d

Solution :

Step 1 of 3 :

Write down the given AP

Here the given AP is 1 , 3 , 5 , 7 , 9 . . .

Step 2 of 3 :

Write down first term and common difference

First term = a = 1

Common Difference = d = 3 - 1 = 2

Step 3 of 3 :

Calculate sum of first 20 terms of AP

We have to calculate sum of first 20 terms of AP 1 , 3 , 5 , 7 , 9 . . .

Number of terms = n = 20

Hence the required sum

= The sum of first 20 terms of AP 1 , 3 , 5 , 7 , 9 . . .

$\displaystyle \sf{ = S_{20} }$

$\displaystyle \sf = \frac{n}{2} \bigg[2a + (n - 1)d \bigg]$

$\displaystyle \sf = \frac{20}{2} \times \bigg[(2 \times 1) + (20 - 1) \times 2 \bigg]$

$\displaystyle \sf = 10 \times \bigg[(2 \times 1) + (19 \times 2) \bigg]$

$\displaystyle \sf = 10 \times (2 + 38)$

$\displaystyle \sf{ = 10 \times 40 }$

$= 400$

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Answer 2

As we know the sum of n terms of an AP is,

$S_n=\frac{n}{2}(2a+(n-1)d)$

Given: AP is 1, 3, 5, 7, 9... and  n  is 20.

Common difference d is,

$d=a_2-a_1\\d=3-1\\d=2$

$S_{20}=\frac{20}{2}(2(1)+(20-1)2)$

$S_{20}=\frac{20}{2}(2+(19)2)$

$S_{20}=10(2+38)$

$S_{20}=10(40)\\S_{20}=400$

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