If the ratio of the sum of n terms of two A.P.s is (3n+1 2n 3 then ratio of their 11th terms is)

Given: Ratio of sum of nth terms of 2 AP’s
To Find: Ratio of their 11th terms

Let us consider 2 AP series AP1 and AP2. Putting n = 1, 2, 3… we get AP1 as 8, 15 22… and AP2 as 31, 35, 39…. So, a1 = 8, d1 = 7 and a2 = 31, d2 = 4 For AP1 S6 = 8 + (11 – 1)7 = 87 For AP2 S6 = 31 + (11 – 1)4 = 81

Required ratio =

8781=2927

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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 = /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 = /2 [2A + (n 1)D] & nth term = An = A + (n 1)D We need to find ratio of their 18th term i.e. (18 1 )/(18 2 ) = ( 18 1 )/( 18 2 ) = (a + (18 1)d)/(A + (18 1)D) = ( + 17 )/(A + 17D) is given that (Sum of n terms of first A )/(Sum of n terms of second A ) = (5n+4)/(9n+6) ( /2[2 +( 1) ])/(( )/2[2 +( 1) ]) = (5n+4)/(9n+6) ( [2 +( 1) ])/( [2 +( 1) ]) = (5n+4)/(9n+6) ( 2(a +(( 1)/2)d))/( 2(A +(( 1)/2)D) ) = (5n+4)/(9n+6) ( (a +(( 1)/2)d))/( (A +(( 1)/2)D) ) = (5n+4)/(9n+6) We have to find ( + 17 )/(A + 17D) Hence, ( 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 (18 1 )/(18 2 ) = 179/321 Hence the ratio of 18th term of 1st AP and 18th term of 2nd AP is 179 : 321

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Let the first terms of the two A.P.'s be a and a'; and their common difference be d and d'.
Now,

\[\frac{S_n}{S_n '} = \frac{\left( 3n + 2 \right)}{\left( 2n + 3 \right)}\]

\[ \Rightarrow \frac{\frac{n}{2}\left[ 2a + \left( n - 1 \right)d \right]}{\frac{n}{2}\left[ 2a' + \left( n - 1 \right)d' \right]} = \frac{\left( 3n + 2 \right)}{\left( 2n + 3 \right)}\]

\[ \Rightarrow \frac{\left[ 2a + \left( n - 1 \right)d \right]}{\left[ 2a' + \left( n - 1 \right)d' \right]} = \frac{\left( 3n + 2 \right)}{\left( 2n + 3 \right)}\]

\[\text { Let }n = 23\]

\[ \Rightarrow \frac{2a + \left( 23 - 1 \right)d}{2a' + \left( 23 - 1 \right)d'} = \frac{3 \times 23 + 2}{2 \times 23 + 3}\]

\[ \Rightarrow \frac{2a + 22d}{2a' + 22d'} = \frac{69 + 2}{46 + 3}\]

\[ \Rightarrow \frac{2\left( a + 11d \right)}{2\left( a' + 11d' \right)} = \frac{71}{49}\]

\[ \therefore \frac{a_{12}}{a_{12'} } = \frac{71}{49}\]

So, the ratio of their 12th terms is 71 : 49.

If the sums of n terms of two A.P,'s are in the ratio 3 n+2:2 n+3, find the ratio of their 12 th terms.

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