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# Hard Algebra 2 problem, anyone can lend me a hand, Thank YOU if you try or help :)

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If {$$a_n$$} is an arithmetic sequence, $$a_1$$= 1, and $$a_{100}$$= 625, evaluate:

$$1\over{\sqrt{a_1}+\sqrt{a_2}}$$+$$1\over{\sqrt{a_2}+\sqrt{a_3}}$$+$$1\over{\sqrt{a_3}+\sqrt{a_4}}$$+...+$$1\over{\sqrt{a_{99}}+\sqrt{a_{100}}}$$

#1
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Note

1/ [sqrt a2 + sqrt a1 ]    if we multiply  top/bottom  by the conjugate we have

[ sqrt  a2 - sqrt a1 ] / [ a2 -a1 ] =   [ sqrt a2 - sqrt 1 ] / d

Also

1/ [sqrt a3 + sqrt a2 ]    if we multiply  top/bottom  by the conjugate we have

[ sqrt  a3 - sqrt a2 ] / [ a3 -a2 ] =   [ sqrt a3 - sqrt a2 ] / d

So we   have    [sqrt a2 - sqrt a1 ] / d  +  [sqrt a3 - sqrt a2 ]  /d + [ sqrt a4 -sqrt a3]/ d   +.....+

[sqrt a99 - sqrt a98 ] /d + [sqrt a100 - sqrt a99] / d

So....eventually we just end up with

[  -sqrt a1  + sqrt a100 ]  /  d  =    [ -1 + 25 ] / d

We can find d by noting that

a1  +  99d  =  a100

1 + 99d  = 625

624  =99d

624/99  = d =208/ 33

So...we have

[ 24 ] / [ 208/33]  = 24 * 33 / 208 =  99 / 26

Apr 7, 2024
#2
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Thanks a lot!