Question

Într o progresie aritmetica avem: a3=9 și a45=15
A8=? ​

Answer 1

a3=9

a45=15

a8= ?

a3=a1+2r

a45=a1+44r

Facem diferenta:

a45-a3=a1+44r-a1-2r <=> a45-a3=42r <=> 42r=6 => r=1/7

a1=?

a3=a1+2r =>a1=a3-2r= 9-2/7= 61/7

a8=a1+7r= 61/7 +7*1/7= 68/7

Answer 2

$a_3 = 9 \\ a_{45} = 15 \\ \\ a_{45} = a_3+42r\Rightarrow 15=9+42r \Rightarrow 42r=</strong><strong>1</strong><strong>5</strong><strong>-</strong><strong>9</strong><strong> \Rightarrow </strong><strong>4</strong><strong>2</strong><strong>r =</strong><strong>6</strong><strong>\</strong><strong>\</strong><strong> </strong><strong>\</strong><strong>R</strong><strong>i</strong><strong>g</strong><strong>h</strong><strong>t</strong><strong>a</strong><strong>r</strong><strong>r</strong><strong>o</strong><strong>w</strong><strong> </strong><strong>r</strong><strong>=</strong><strong> </strong><strong>\</strong><strong>d</strong><strong>f</strong><strong>r</strong><strong>a</strong><strong>c</strong><strong>{</strong><strong>1</strong><strong>}</strong><strong>{</strong><strong>7</strong><strong>}</strong><strong> </strong><strong>\</strong><strong>\</strong><strong> </strong><strong>\</strong><strong>\</strong><strong>a_8 = a_3+5r = 9+\dfrac{5}{7} = \boxed{\dfrac{68}{7}}$