Question

Em uma P.A. de vinte e cinco termos, a soma de todos os termos é 2525. Calcule a soma dos últimos cinco termos, sabendo que a soma dos cinco primeiros é 105

Answer 1

$\mathrm{PA=\{a_1,a_2,a_3,\dots,a_{25}\}\ \to\ n=25}\\\\ \mathrm{S_n=2525\ \to\ n\bigg[\dfrac{a_1+a_n}{2}\bigg]=2525\ \to\ 25\bigg[\dfrac{a_1+a_{25}}{2}\bigg]=2525}\\\\ \mathrm{\dfrac{a_1+a_1+24r}{2}=\dfrac{2525}{25}\ \to\ a_1+12r=101\ \to\ \boxed{\mathrm{a_{13}=21}}}\\\\ \mathrm{S_{5}=105\ \to\ 5\bigg[\dfrac{a_1+a_5}{2}\bigg]=105\ \to\ \dfrac{a_1+a_1+4r}{2}=\dfrac{105}{5}}\\\\ \mathrm{\dfrac{2a_1+4r}{2}=21\ \to\ a_1+2r=21\ \to\ \boxed{\mathrm{a_3=21}}}\\\\ \mathrm{a_n=a_k+(n-k)r\ \to\ a_{13}=a_3+(13-3)r}\\\\ \mathrm{101=21+10r\ \to\ 10r=80\ \to\ \boxed{\mathrm{r=8}}}\\\\ \mathrm{a_{21}=a_3+(21-3).8\ \to\ a_{21}=21+144\ \to\ \boxed{\mathrm{a_{21}=165}}}\\\\ \mathrm{a_{25}=a_3+(25-3).8\ \to\ a_{25}=21+176\ \to\ \boxed{\mathrm{a_{25}=197}}}\\\\ \mathrm{S=5\bigg[\dfrac{a_{21}+a_{25}}{2}\bigg]\ \to\ S=5\bigg[\dfrac{165+197}{2}\bigg]\ \to\ S=5\bigg[\dfrac{362}{2}\bigg]}\\\\ \mathrm{S=5[181]\ \to\ \boxed{\boxed{\mathbf{S=905}}}}$