If the ratio of the sum of n terms of two A.P | vedclass

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(C) Let $S_n$ and $S'_n$ be the sums of $n$ terms of two $A.P.s$ with first terms $a, a'$ and common differences $d, d'$ respectively.Given $\frac{S_n

Vedclass · Accepted Answer

$4:3$

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(C) Let $S_n$ and $S'_n$ be the sums of $n$ terms of two $A.P.s$ with first terms $a, a'$ and common differences $d, d'$ respectively.Given $\frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a + (n - 1)d]}{\frac{n}{2}[2a' + (n - 1)d']} = \frac{7n + 1}{4n + 27}$.This simplifies to $\frac{a + \frac{n-1}{2}d}{a' + \frac{n-1}{2}d'} = \frac{7n + 1}{4n + 27}$.To find the ratio of the $11^{th}$ terms,we need the expression $\frac{a + 10d}{a' + 10d'}$.Comparing $\frac{n-1}{2} = 10$,we get $n-1 = 20$,so $n = 21$.Substituting $n = 21$ in the ratio:$\frac{T_{11}}{T'_{11}} = \frac{7(21) + 1}{4(21) + 27} = \frac{147 + 1}{84 + 27} = \frac{148}{111}$.Dividing both by $37$,we get $\frac{148 \div 37}{111 \div 37} = \frac{4}{3}$.Thus,the ratio is $4:3$.

If the ratio of the sum of $n$ terms of two $A.P.s$ is $(7n + 1):(4n + 27)$,then the ratio of their $11^{th}$ terms is:

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