Ex 9.3, 24 - Class 11

Transcript

We need to show ratio of the sum of n terms of GP & sum of terms from (n + 1)th to 2nth term i.e. we need to calculate (𝑆𝑢𝑚 𝑜𝑓 𝑛 𝑡𝑒𝑟𝑚𝑠)/(𝑆𝑢𝑚 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠 𝑓𝑟𝑜𝑚 (𝑛 + 1)𝑡ℎ 𝑡𝑜 (2𝑛)𝑡ℎ 𝑡𝑒𝑟𝑚𝑠) Putting values from (1) & (2) = (( a(1 −〖 𝑟〗^𝑛))/((1 − r)))/(( a)/((1 − r))(〖 𝑟〗^𝑛−〖 𝑟〗^2𝑛)" " ) = (( a(1 −〖 𝑟〗^𝑛))/((1 − r)))/(( a)/((1 − r) ) 〖 𝑟〗^𝑛 (1 −〖 𝑟〗^𝑛)" " ) = (( a(1 −〖 𝑟〗^𝑛))/((1 − r)))/𝑟𝑛(( a(1 −〖 𝑟〗^𝑛))/((1 − r))) = ( a(1 − 𝑟^𝑛))/(rn(1 −〖 𝑟〗^𝑛)) × ( 1 − r)/(1 − r) Ex 9.3, 24 Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n+1)th to (2n)th term is ( 1)/rn Ex 9.3, 24 Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n+1)th to (2n)th term is ( 1)/rn We need to show ratio of the sum of n terms of GP & sum of terms from (n + 1)th to 2nth term i.e. we need to calculate (𝑆𝑢𝑚 𝑜𝑓 𝑛 𝑡𝑒𝑟𝑚𝑠)/(𝑆𝑢𝑚 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠 𝑓𝑟𝑜𝑚 (𝑛 + 1)𝑡ℎ 𝑡𝑜 (2𝑛)𝑡ℎ 𝑡𝑒𝑟𝑚𝑠) Putting values from (1) & (2) = (( a(1 −〖 𝑟〗^𝑛))/((1 − r)))/(( a)/((1 − r))(〖 𝑟〗^𝑛−〖 𝑟〗^2𝑛)" " ) = (( a(1 −〖 𝑟〗^𝑛))/((1 − r)))/(( a)/((1 − r) ) 〖 𝑟〗^𝑛 (1 −〖 𝑟〗^𝑛)" " ) = (( a(1 −〖 𝑟〗^𝑛))/((1 − r)))/𝑟𝑛(( a(1 −〖 𝑟〗^𝑛))/((1 − r))) = ( a(1 − 𝑟^𝑛))/(rn(1 −〖 𝑟〗^𝑛)) × ( 1 − r)/(1 − r) = 1/𝑟𝑛