Question

Calcule a soma dos primeiros:
a) 15 termos da PA(5,9,13...)
b) 36 termos da PA (17,11,5...)

Answer 1

1° vc  terá que descobrir qual é o último termo de cada PA:

a) 15 termos da PA(5,9,13...)

$a_{n} = a_{1} +(n-1)*r\\ a_{15} =5+(15-1)*4\to\\ a_{15} =5+14*4\to \\a_{15} =5+56\to \\a_{15} =61$

Agora passamos a soma dos termos:

$S_{n} = \frac{( a_{1} + a_{n} )*n}{2} \to \\\\ S_{15} = \frac{( 5 + 61 )*15}{2} \to \\\\ S_{15} = \frac{(66 )*15}{2} \to \\\\ S_{15} = \frac{990}{2} \to \\\\ S_{15} = 495$







b) 36 termos da PA (17,11,5...)

$a_{n} = a_{1} +(n-1)*r\\ a_{36} =17+(36-1)*(-6)\to\\ a_{36} =17+(35)*(-6)\to\\ a_{36} =17-210\to\\ a_{36} =-193$

Agora passamos a soam dos termos:


$S_{n} = \frac{( a_{1} + a_{n} )*n}{2} \to \\\\ S_{36} = \frac{(17 + (-193) )*36}{2} \to \\\\ S_{36} = \frac{(17-193 )*15}{2} \to \\\\ S_{36} = \frac{(-176)*36}{2} \to \\\\ S_{36} = \frac{6336}{2} \to \\\\ S_{36} = 3168$


Prontinho, espero ter ajudado...