The sum of the 3rd and 7th term of an a. p is 6, the product is 8, find the sum of the first 16 terms of the sequence

For starter, recall that an arithmetic progression is given recursively by

We can solve this explicitly for in terms of :

and so on, down to

More generally, we can express in terms of an arbitrary term :

(notice how the index of and the coefficient of on the right side add to )

Let be the -th term of this particular progression. We're given

from which it follows that

If , it follows that

Otherwise, if , then

The sum of the first terms of an arithmetic progression is

If , then

so that

Otherwise, if , then

so that

what "following statement"?

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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