Question

Am o nelamurire in cazul a doua exercitii.Amandoua sunt cu progresii aritmetice.
1.a2=4
a5=10
Suma primilor 10 termeni(S10)=???
Cum as putea sa calculez in cazul in care nu cunosc primul termen.
2.a6+a9+a12+a15=20
S20=???
Aceeasi nelamurire.Nu stiu cum sa calculez daca nu cunosc primul termen.

Answer 1

1.
Formula termenului general:
$a_n=a_1+(n-1)r$

[tex]a_2=a_1+r\\ a_5=a_1+4r\\\\ \left \{ {{a_1+r=4} \atop {a_1+4r=10}} \right.\rightarrow 3r=6 \rightarrow r=2\\\\ a_1=4-r=2[/tex]

$S_{10}= \frac{10(a_1+a_{10})}{2} = \frac{10(2a_1+9r)}{2}=110$

2.

O sa grupam termenii astfel:

[tex](a_6+a_{15})+(a_{9}+a_{12})=20\\ (a_1+5r+a_1+14r)+(a_1+8r+a_1+11r)=20\\ (2a_1+19r)+(2a_1+19r)=20\\ 2(a_1+a_1+19r)=20\\ a_1+a_1+19r=10\\ a_1+19r=a_{20}\rightarrow a_1+a_{20}=10[/tex]

$S_{20}= \frac{20(a_1+a_{20})}{2} =100$

Answer 2

1) Fie r- rația progresiei.

a₅ = a₂+3r ⇒ r = (a₅ - a₂)/3 = (10 - 4)/3 = 6/3 = 2

a₁ = a₂-r = 4-2 = 2

a₁₀ = a₁+9r = 2+9·2 =20

S₁₀ = (a₁+ a₁₀ )·10/2 = (2+ 20) ·5  = 22··5 = 110

2) Fie r - rația progresiei.

[tex]\it a_6+a_9+a_{12}+a_{15} =20 \Leftrightarrow a_6+a_6+3r+a_6+6r+a_6+9r =20 \Leftrightarrow \\\;\\ \Leftrightarrow 4a_6+18r = 20|_{:2} \Leftrightarrow 2a_6+9r = 10 \Leftrightarrow a_6+a_6+9r=10\Leftrightarrow \\\;\\ \Leftrightarrow a_6+a_{15}=10 \ \ \ \ \ (*)[/tex]

[tex]\it S_{20} =\dfrac{(a_1+a_{20})\cdot20}{2} = (a_6-5r+a_{15}+5r)\cdot10 =(a_6+a_{15}) \cdot10 \stackrel{(*)}{=} \\\;\\ \\\;\\ = 10\cdot10=100. [/tex]