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Determinanti ratia progresiei aritmetice a1+a2+a3+a4=14 si a1=2

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[tex] a_{1} [/tex]=2
[tex]a_{2} [/tex]=3
[tex]a_{3} [/tex]=4
[tex] a_{4} [/tex]=5
S=[tex] \frac{ {n} ( a_{n}+ a_{1} ) }{2} = \frac{ {n} ( a_{4}+ a_{1} ) }{2} =\frac{ 4 ( 5+2) }{2} =\frac{ {4} x7 }{2} = \frac{28}{2} =14[/tex]
[tex]r= a_{n} - a_{n-1} =5-4=1[/tex]

r=1
a2=a1+r
a3=a1+2r
a4=a1+3r
a1+a1+r+a1+2r+a1+3r=14
4*2+6r=14
8+6r=14
6r=6
r=1

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