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Question

# The sum of n terms of two arithmetic progressions are in the ratio (3n+8):(7n+15). Find the ratio of their 12th terms.

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Solution

## Simplication of ratio of their nth terms. Let (A.P.)1→a1 and d1 are the first term and common difference respectively (A.P.)2→a2 and d2 are the first term and common difference respectively According to given condition, we have Sum of n terms of A.P.1Sum of n terms of A.P.2=(3n+8)(7n+15) ⇒n2[2a1+(n−1)d1]n2[2a2+(n−1)d2]=3n+87n+15 ⇒2a1+(n−1)d12a2+(n−1)d2=3n+87n+15 ⇒2[a1+n−12d1]2[a2+n−12d2]=3n+87n+15 ⇒a1+(n−1)2d1a2+(n−1)2d2=3n+87n+15⋯(i) Finding the ratio of their 12th terms. Substitute the coefficient of common difference to 11. ∴n−12=11 ⇒n−1=22 ⇒n=23 substitute this in equation (i) ∴a1+11d1a2+11d2=3×23+87×23+15 a1+11d1a2+11d2=77176 ∴12th term of A.P.112th term of A.P.2=716 Final answer: Hence, the required ratio is 7:16.

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