Example 6 - Sum of n terms of two APs are in ratio (3n + 8) - Arithmetic Progression (AP): Formulae based

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Example 6 The sum of n terms of two arithmetic progressions are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 12th terms. There are 2 AP’s with different first term and common difference For the first AP Let first term be a common difference be d Sum of n term = Sn = 𝑛/2 (2a + (n – 1)d) & nth term = an = a + (n – 1)d Similarly for second AP Let first term = A common difference = D Sn = 𝑛/2 (2A + (n – 1)D) & nth term = An = A + (n – 1)D We need to find ratio of 12th term i.e. (𝑎12 𝑜𝑓 𝑓𝑖𝑟𝑠𝑡 𝐴𝑃 )/(𝐴12 𝑜𝑓 𝑠𝑒𝑐𝑜𝑛𝑑 𝐴𝑃) = (a +(12 − 1)d)/(A +(12 − 1)D) = (a + 11d)/(A + 11𝐷) It is given that (𝑆𝑢𝑚 𝑜𝑓 𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 1𝑠𝑡 𝐴𝑃 )/(𝑆𝑢𝑚 𝑜𝑓 𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 2𝑛𝑑 𝐴𝑃) = (3𝑛 + 8)/(7𝑛 + 15 ) (𝑛/2[2𝑎+(𝑛 − 1)𝑑])/((𝑛 )/2[2𝐴+(𝑛 − 1)𝐷]) = (3n+8)/(7n+15) ( [2a + (n − 1)d])/( [2A + (n − 1)D]) = (3n+8)/(7n+15) ( 2(a +((𝑛 −1)/2)d))/( 2(A +((𝑛 −1)/2)D) ) = (3n+8)/(7n+15) ( (a +((𝑛 −1)/2)d))/( (A +((𝑛 −1)/2)D) ) = (3n+8)/(7n+15) We need to find (a + 11d)/(A + 11D) Hence (𝑛 − 1)/2 = 11 n – 1 = 22 n = 23 Putting n = 23 in (1) ("a + (" (23 − 1)/2 ")d" )/(𝐴+"(" (23 − 1)/2 ")" 𝐷) = (3 × 23 + 8)/(7 × 23 +15) ("a + (" 22/2 ")d" )/(A+"(" 22/2 ")" D) = (69 + 8)/(161 +15) (a+11d)/(A + 11𝐷) = 77/176 (a+11d)/(A + 11𝐷) = 7/16 Hence ratio of their 12th term is 7/16 i.e. 7 : 16

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