Question

Find the sum of 1+3+5+...+to 40 terms​

Answer 1

The required sum = 1600

Given :

The progression 1 + 3 + 5 + . . . + upto 40 terms

To find :

The sum

Concept :

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 given progression

Here the given progression is

1 + 3 + 5 + . . . + upto 40 terms

This is an arithmetic progression

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 the sum

Number of terms = 40

Hence the required sum

$\displaystyle \sf = \frac{40}{2} \times \bigg[(2 \times 1)+ (40 - 1) \times 2 \bigg]$

$\displaystyle \sf = 20 \times \bigg[2+ (39 \times 2) \bigg]$

$\displaystyle \sf = 20 \times \bigg[2+ 78 \bigg]$

$\displaystyle \sf = 20 \times 80$

$\displaystyle \sf{ = 1600 }$

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