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# The ratio of the $ 11^{\text {th }} $ term to the $ 18^{\text {th }} $ term of an AP is $ 2: 3 $. Find the ratio of the $ 5^{\text {th }} $ term to the 21 term

Given:

The ratio of the \( 11^{\text {th }} \) term to the \( 18^{\text {th }} \) term of an AP is \( 2: 3 \).

To do:

We have to find the ratio of the \( 5^{\text {th }} \) term to the 21 " term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.

Solution:

Let $a$ be the first term and $d$ be the common difference of the AP.

This implies,

$a_{11}: a_{18} =2: 3$

$\frac{a+10 d}{a+17 d}=\frac{2}{3}$

$3 a+30 d=2 a+34 d$

$3a-2a=34 d-30d$

$a=4d$........(i)

$a_5=a+(5-1)d$

$=a+4d$

$a_{21}=a+(21-1)d$

$=a+20d$

This implies,

$a_5:a_{21}=(a+4d):(a+20d)$

$=\frac{4d+4d}{4d+20d}$

$=\frac{8d}{24d}$

$=\frac{1}{3}$

Sum of the first five terms $S_{5}=\frac{5}{2}[2 a+(5-1) d]$

$=\frac{5}{2}[2(4 d)+4 d]$

$=\frac{5}{2}(8 d+4 d)$

$=\frac{5}{2}(12 d)$

$=30 d$

Sum of the first 21 terms $S_{21}=\frac{21}{2}[2 a+(21-1) d]$

$=\frac{21}{2}[2(4 d)+20 d]$ [From (i)]

$=\frac{21}{2}(28 d)$

$=294 d$

Therefore,

The ratio of the sum of the first five terms to the sum of the first 21 terms is,

$S_5:S_{21}=30 d: 294 d$

$=5: 49$

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