Question

a soma dos termos de uma p.g (5,50...,500000) é?

Answer 1

A P.G (Progressão geométrica) é:

$\mathsf{PG:\left(5,50,...,50000\right)}$

Onde iremos inicialmente calcular o valor da razão:

$\mathsf{q=\dfrac{a_n}{a_{n-1}}}\\\\\\\mathsf{q=\dfrac{a_2}{a_1}}\\\\\\\mathsf{q=\dfrac{50}{5}}\\\\\\\mathsf{q=10}$

Agora vamos determinar quantos termos tem essa P.G:

$\mathsf{a_n=a_k\cdot q^{n-k}}\\\mathsf{a_n=a_1\cdot q^{n-1}}\\\mathsf{500000=5\cdot 10^{n-1}}\\\mathsf{5\cdot 10^5=5\cdot 10^{n-1}}\\\mathsf{10^5=10^{n-1}}\\\mathsf{n-1=5}\\\mathsf{n=5+1}\\\mathsf{n=6}$

Substituindo na formula da soma de uma P.G finita:

$\mathsf{S_n=a_1\cdot \dfrac{q^n-1}{q-1}}\\\\\\\mathsf{S_6=5\cdot \dfrac{10^6-1}{10-1}}\\\\\\\mathsf{S_6=5\cdot \dfrac{10^6-1}{9}}\\\\\\\mathsf{S_6=5\cdot \dfrac{1000000-1}{9}}\\\\\\\mathsf{S_6=5\cdot \dfrac{999999}{9}}\\\\\\\mathsf{S_6=5\cdot 111111}\\\\\\\boxed{\boxed{\mathbf{S_6=555555}}}$