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

Explanation

According to question

`(n/2[2a1​+(n′−1)d1​]
)/(n/2[2a2​+(n′−1)d2​])​=(2n+3​)/(3n+2)`


Put n′=10 for the ratio of 10th terms

`⇒(a1​+9d1​​ )/(a2​+9d2​)=(2×19+3​)/(3×19+2)`


[putting`2n−1​=9=>n=19]`
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