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The sum of n terms of two arithmetic progressions are in the ...

Then, (n/2(2a+(n-1)d1))/(n/2(2b+(n-1)d2)) = (3n+8)/(7n+15) =>(2a+(n-1)d1)/(2b+(n-1)d2) =(k(3n+8))/(k(7n+15)) Comparing numerators for both sides, ...

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