Question

insira quartoze meios aritméticos entre 102 e 147.​

Answer 1

$\large\boxed{\begin{array}{l}\sf Inserir\,meios\,aritm\acute eticos\,entre\,dois\,n\acute umeros\\\sf \acute E\,formar\,uma\, PA\,contendo\,esses\,mesmos\,n\acute umeros.\\\sf Neste\,caso\,queremos\,formar\,uma\,PA\,inserindo\\\sf 14\,termos\,entre\,102\,e\,147.\\\sf Ao\,todo\,teremos\,16\,elementos\\\sf e\,para\,formar\,esta\,PA\,precisamos\,descobrir\,a\,raz\tilde ao.\end{array}}$

$\small\boxed{\begin{array}{l}\sf PA~(102,\blacktriangle,\blacktriangle,\blacktriangle,\blacktriangle,\blacktriangle,\blacktriangle,\blacktriangle,\blacktriangle,\blacktriangle,\blacktriangle,\blacktriangle,\blacktriangle,\blacktriangle,\blacktriangle,147)\\\sf a_1=102~~a_{16}=147.\\\sf a_{16}=a_1+15r\\\sf 147=102+15r\\\sf 15r=147-102\\\sf 15r=45\\\sf r=\dfrac{45}{15}\\\\\sf r=3\\\sf (102,\boldsymbol{105,108,111,114,117,120,123,126,129,132,135,138}\\\sf\boldsymbol{141,144},147)\end{array}}$