write an A. P. in which a=10 and d is any natural number. find the sum of first 20 terms using formula.​

The Sum of first 20 terms of required AP is 2100.

Justification:

Let us suppose d is equal to 10,

then,

Structure of AP => a, a + d, a + 2d

AP => 10, 20, 30, 40...

Now,

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

Sn = 20/2(2 × 10 + (20 - 1)10)

Sn = 10(20 + 190)

Sn = 10 × 210

Sn = 2100.

Here, a = 10 . Let d= 5. The A.P. is 10,15,20..

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

Hence, the sum of first 20 terms is 1150.

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step-by-step explanation:

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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