Question

The ratio of the sum of n terms of an AP is 2n-5 : n+7. Find the ratio of their 15 terms. ​

Answer 1

The ratio of their 15 terms 53 : 36

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$\orange{\bigstar}$  Given  $\green{\bigstar}$

The ratio of the sum of n terms of an AP is 2n-5 : n+7

$\orange{\bigstar}$  To Find  $\green{\bigstar}$

The ratio of their 15 terms

$\orange{\bigstar}$  Key Points  $\green{\bigstar}$

$\rm nth\ term\ of\ an\ AP,\\\\ \pink{\bigstar}\ \; \bf a_n=a+(n-1)d\\\\\rm Sum\ of\ n\ terms\ in\ an\ AP,\\\\\pink{\bigstar}\ \; \bf S_n=\dfrac{n}{2}(2a+(n-1)d)$

$\orange{\bigstar}$  Solution  $\green{\bigstar}$

Let ,

For 1st AP :

nth term = aₙ

First term = a

Common difference = d

For 2nd AP :

nth term = Aₙ

First term = A

Common difference = D

So , A/c , " The ratio of the sum of n terms of an AP is 2n-5 : n+7 "

$\to \rm \dfrac{\frac{n}{2}(2a+(n-1)d)}{\frac{n}{2}(2A+(n-1)D) }=\dfrac{2n-5}{n+7}\\\\\to \rm \dfrac{2(a+(\frac{n-1}{2})d)}{2(A+(\frac{n-1}{2})D)}=\dfrac{2n-5}{n+7}\\\\\to \bf \dfrac{a+(\frac{n-1}{2})d}{A+(\frac{n-1}{2})D}=\dfrac{2n-5}{n+7}...(1)$

Here , we need to find the ratio of  $\bf \dfrac{a_{15}}{A_{15}}= \dfrac{a+14d}{A+14D}\ \; \red{\bigstar}$ ,

So ,

$\to \rm \dfrac{n-1}{2}=14\\\\\to \rm n-1=28\\\\\to \bf n=29\ \; \pink{\bigstar}$

So , sub. n value in (1) ,

$\to \rm \dfrac{a+(\frac{29-1}{2})d}{A+(\frac{29-1}{2})D}=\dfrac{2(29)-5}{(29)+7}\\\\\to \bf \dfrac{a+14d}{A+14D}=\dfrac{53}{36}\\\\\to \bf \dfrac{a_{15}}{A_{15}}=\dfrac{53}{36}\ \; \pink{\bigstar}$

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