Ex 9.2, 9 - Class 11

Transcript

Ex 9.2 , 9 The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms. There are two AP with different first term and common difference For the first AP Let first term be = a Common difference = d Sum of n terms = Sn = 𝑛/2 [2a + (n – 1)d] & nth term = an = a + (n – 1)d For the second AP Let first term be = A common difference = D Sum of n terms = Sn = 𝑛/2 [2A + (n – 1)D] & nth term = An = A + (n – 1)D We need to find ratio of their 18th term i.e. (18𝑡ℎ 𝑡𝑒𝑟𝑚 𝑜𝑓 1𝑠𝑡 𝐴𝑃 )/(18𝑡ℎ 𝑡𝑒𝑟𝑚 𝑜𝑓 2𝑛𝑑 𝐴𝑃) = (𝑎18 𝑜𝑓 1𝑠𝑡 𝐴𝑃 )/(𝐴18 𝑜𝑓 2𝑛𝑑 𝐴𝑃) = (a + (18 − 1)d)/(A + (18 − 1)D) = (𝑎 + 17𝑑)/(A + 17D) is given that (Sum of n terms of first A𝑃)/(Sum of n terms of second A𝑃) = (5n+4)/(9n+6) (𝑛/2[2𝑎+(𝑛 − 1)𝑑])/((𝑛 )/2[2𝐴+(𝑛 − 1)𝐷]) = (5n+4)/(9n+6) ( [2𝑎+(𝑛 − 1)𝑑])/( [2𝐴+(𝑛 − 1)𝐷]) = (5n+4)/(9n+6) ( 2(a +((𝑛 −1)/2)d))/( 2(A +((𝑛 −1)/2)D) ) = (5n+4)/(9n+6) ( (a +((𝑛 −1)/2)d))/( (A +((𝑛 −1)/2)D) ) = (5n+4)/(9n+6) We have to find (𝑎 + 17𝑑)/(A + 17D) Hence, (𝑛 −1)/2 = 17 n – 1 = 17 × 2 n – 1 = 34 n = 34 + 1 n = 35 Putting n = 35 in (1) ( (a +((35 −1)/2)d))/( (A +((35 −1)/2)D) ) " "= (5(35)+4)/(9(35)+6) ( (a +(34/2)d))/( (A +(34/2)D) )= (175 + 4)/(315 + 6) ([a + 17d])/( [A + 17D]) = 179/321 Therefore (18𝑡ℎ 𝑡𝑒𝑟𝑚 𝑜𝑓 1𝑠𝑡 𝐴𝑃 )/(18𝑡ℎ 𝑡𝑒𝑟𝑚 𝑜𝑓 2𝑛𝑑 𝐴𝑃) = 179/321 Hence the ratio of 18th term of 1st AP and 18th term of 2nd AP is 179 : 321