Question

41+42+43+44+45+....80​

Answer 1

Answer:

45 + 42 = 462 + 43 = 505 + 45 = 550 + 80 = 630

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Answer 2

We need to find the sum of numbers from 41 to 80.

Let's find the solution through the concept of arithmetic progression.

Given series is

$41 + 42 + 43 + ......... + 80$

It is in AP.

The first term is

$a = 41$

The common difference

$d = a1 - a0 \\ = 42 - 41 \\ = 1$

Therefore common difference is

$1$

Number of terms that are present between

$41 \: and \: 80 \: is \: 40$

There are

$40$

number of terms in the given series.

We know the sum of n terms formula in AP.

$sn = n \div 2(2a + (n - 1)d) \\ s40 = 40 \div 2(2 \times 41 + (40 - 1) \times 1) \\ = 20(82 + 39) \\ = 20(121) \\ s40= 2420$

Therefore, the given series results in

$2420$

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