


Ex 9.3
Ex 9.3, 2
Ex 9.3, 3 Important
Ex 9.3, 4
Ex 9.3, 5 (a)
Ex 9.3, 5 (b) Important
Ex 9.3, 5 (c)
Ex 9.3, 6
Ex 9.3, 7 Important
Ex 9.3, 8
Ex 9.3, 9 Important
Ex 9.3, 10
Ex 9.3, 11 Important
Ex 9.3, 12
Ex 9.3, 13
Ex 9.3, 14 Important
Ex 9.3, 15
Ex 9.3, 16 Important
Ex 9.3, 17 Important
Ex 9.3, 18 Important
Ex 9.3, 19
Ex 9.3, 20
Ex 9.3, 21
Ex 9.3, 22 Important
Ex 9.3, 23 Important
Ex 9.3, 24
Ex 9.3, 25
Ex 9.3, 26 Important
Ex 9.3, 27 Important You are here
Ex 9.3, 28
Ex 9.3, 29 Important
Ex 9.3, 30 Important
Ex 9.3, 31
Ex 9.3, 32 Important
Ex 9.3
Last updated at May 29, 2018 by Teachoo
Ex9.3, 27 Find the value of n so that (𝑎^(𝑛 + 1) +𝑏^(𝑛 + 1))/(𝑎^(𝑛 ) +𝑏^𝑛 ) may be the geometric mean between a and b. We know that geometric mean between a & b is a & b = √ab It is given that G.M. between a & b = (𝑎^(𝑛 + 1) +𝑏^(𝑛 + 1))/(𝑎^(𝑛 ) +𝑏^(𝑛 ) ) √ab = (𝑎^(𝑛 + 1) +𝑏^(𝑛 + 1))/(𝑎^(𝑛 ) +𝑏^(𝑛 ) ) 〖"(ab)" 〗^(1/2) = (𝑎^(𝑛 + 1) +𝑏^(𝑛 + 1))/(𝑎^(𝑛 ) +𝑏^(𝑛 ) ) 〖"(ab)" 〗^(1/2) (an +bn) = an + 1 + bn + 1 〖"a" 〗^(1/2) 𝑏^(1/2) (an +bn) = an + 1 + bn + 1 〖"a" 〗^(1/2) an 𝑏^(1/2) + 〖"a" 〗^(1/2) bn 𝑏^(1/2) = an + 1 + bn + 1 𝑎^(1/2 + 𝑛 ) 𝑏^(1/2) + 〖"a" 〗^(1/2) 𝑏^(1/2 + 𝑛 )= an + 1 + bn + 1 𝑎^(1/2 + 𝑛 ) 𝑏^(1/2) – an + 1 = bn + 1 – 〖"a" 〗^(1/2) 𝑏^(1/2 + 𝑛 ) 𝑎^(1/2 + 𝑛 ) 𝑏^(1/2) – 𝑎^(𝑛 + 1/2 + 1/2) = 𝑏^(𝑛 + 1/2 + 1/2) – 𝑎^(1/2) 𝑏^(1/2 + 𝑛 ) 𝑎^(1/2 + 𝑛 ) [𝑏^(1/2) – 𝑎^(1/2)] = 𝑏^(𝑛 + 1/2 ) [𝑏^(1/2) – 𝑎^(1/2)] 𝑎^(1/2 + 𝑛 )= 𝑏^(𝑛 + (1 )/2 "[" 𝑏^(1/2) " − " 𝑎^(1/2) "] " )/(𝑏^(1/2) " − " 𝑎^(1/2) ) 𝑎^(1/2 + 𝑛 )= 𝑏^(𝑛 +1/2) (𝑎/𝑏)^(1/2 + 𝑛) = 1 (𝑎/𝑏)^(1/2 + 𝑛)= (𝑎/𝑏)^0 Comparing powers 1/2 + n = 0 n = – 1/2 Hence value of n is - 1/2