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Worked Examples · Example 13

Q.The sums of nn terms of two A.P.s are in the ratio (3n+8):(7n+15)(3n+8):(7n+15). Find the ratio of their 12th terms.

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Since S2n−1=(2n−1)anS_{2n-1}=(2n-1)a_n (the sum of an odd number of A.P. terms equals that count times the middle/nnth term), substituting 2n−1=232n-1=23 into the given sum-ratio directly gives the ratio of 12th terms, 7:16.

For an A.P., the sum of the first (2n−1)(2n-1) terms equals (2n−1)(2n-1) times the nnth term (the middle term of that block):

S2n−1=(2n−1) an⟹an=S2n−12n−1S_{2n-1} = (2n-1)\,a_n \quad\Longrightarrow\quad a_n = \frac{S_{2n-1}}{2n-1}

Given: Sn(AP1)Sn(AP2)=3n+87n+15\dfrac{S_n(\text{AP}_1)}{S_n(\text{AP}_2)} = \dfrac{3n+8}{7n+15} for all nn.

  1. We want the ratio of the 12th terms, a12(AP1)a12(AP2)\dfrac{a_{12}(\text{AP}_1)}{a_{12}(\text{AP}_2)}.
  2. By the formula above, a12=S2323a_{12} = \dfrac{S_{23}}{23} for each A.P. (taking 2n−1=23⇒n=122n-1=23 \Rightarrow n=12).
  3. So a12(AP1)a12(AP2)=S23(AP1)/23S23(AP2)/23=S23(AP1)S23(AP2)\dfrac{a_{12}(\text{AP}_1)}{a_{12}(\text{AP}_2)} = \dfrac{S_{23}(\text{AP}_1)/23}{S_{23}(\text{AP}_2)/23} = \dfrac{S_{23}(\text{AP}_1)}{S_{23}(\text{AP}_2)}.
  4. Use the given ratio formula with n=23n=23: S23(AP1)S23(AP2)=3(23)+87(23)+15\dfrac{S_{23}(\text{AP}_1)}{S_{23}(\text{AP}_2)} = \dfrac{3(23)+8}{7(23)+15}. …

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