The sum of n terms of two arithmetic progress | vedclass

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(A) Let $a_1, a_2$ and $d_1, d_2$ be the first terms and common differences of the first and second arithmetic progressions,respectively.Given the rat

Vedclass · Accepted Answer

$7 : 16$

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(A) Let $a_1, a_2$ and $d_1, d_2$ be the first terms and common differences of the first and second arithmetic progressions,respectively.Given the ratio of the sum of $n$ terms:$\frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{3n + 8}{7n + 15}$$\frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{3n + 8}{7n + 15}$  --- $(1)$We need the ratio of the $12^{	ext{th}}$ terms,which is $\frac{a_1 + 11d_1}{a_2 + 11d_2}$.To match the form in equation $(1)$,we multiply the numerator and denominator by $2$:$\frac{2a_1 + 22d_1}{2a_2 + 22d_2} = \frac{a_1 + 11d_1}{a_2 + 11d_2}$Comparing $(n-1) = 22$,we get $n = 23$.Substituting $n = 23$ in equation $(1)$:$\frac{a_1 + 11d_1}{a_2 + 11d_2} = \frac{3(23) + 8}{7(23) + 15} = \frac{69 + 8}{161 + 15} = \frac{77}{176}$Dividing both by $11$,we get $\frac{7}{16}$.Thus,the required ratio is $7 : 16$.

The sum of $n$ terms of two arithmetic progressions are in the ratio $(3n + 8) : (7n + 15)$. Find the ratio of their $12^{\text{th}}$ terms.

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