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Answered on 01 Mar Learn Arithmetic Progression

Kalaiselvi

Online Mathematics tutor with 4 years experience(Online Classes for 10th to 12th)

Harmonic progression is defined as series real number which is calculated by taking reciprocals of the arithmetic progression.
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Answered on 16 Apr Learn Arithmetic Progression

To show that the above equation read more

To show that the above equation $To&space;show&space;that&space;$$&space;(a&space;-&space;b)^2&space;$$,&space;$$&space;(a^2&space;+&space;b^2)&space;$$,&space;and&space;$$&space;(a&space;+&space;b)^2&space;$$&space;are&space;in&space;arithmetic&space;progression&space;(AP),&space;we&space;need&space;to&space;demonstrate&space;that&space;the&space;differences&space;between&space;consecutive&space;terms&space;are&space;constant.&space;First,&space;let's&space;find&space;the&space;expressions&space;for&space;these&space;terms:&space;1.&space;$$&space;(a&space;-&space;b)^2&space;$$&space;$&space;(a&space;-&space;b)^2&space;=&space;a^2&space;-&space;2ab&space;+&space;b^2&space;$&space;2.&space;$$&space;(a^2&space;+&space;b^2)&space;$$&space;3.&space;$$&space;(a&space;+&space;b)^2&space;$$&space;$&space;(a&space;+&space;b)^2&space;=&space;a^2&space;+&space;2ab&space;+&space;b^2&space;$&space;Now,&space;let's&space;find&space;the&space;differences&space;between&space;consecutive&space;terms:&space;1.&space;Difference&space;between&space;$$&space;(a&space;-&space;b)^2&space;$$&space;and&space;$$&space;(a^2&space;+&space;b^2)&space;$$:&space;$&space;(a^2&space;+&space;b^2)&space;-&space;(a^2&space;-&space;2ab&space;+&space;b^2)&space;=&space;a^2&space;+&space;b^2&space;-&space;a^2&space;+&space;2ab&space;-&space;b^2&space;=&space;2ab&space;$&space;2.&space;Difference&space;between&space;$$&space;(a^2&space;+&space;b^2)&space;$$&space;and&space;$$&space;(a&space;+&space;b)^2&space;$$:&space;$&space;(a&space;+&space;b)^2&space;-&space;(a^2&space;+&space;b^2)&space;=&space;a^2&space;+&space;2ab&space;+&space;b^2&space;-&space;a^2&space;-&space;b^2&space;=&space;2ab&space;$&space;As&space;we&space;can&space;see,&space;the&space;differences&space;between&space;consecutive&space;terms&space;are&space;both&space;$$&space;2ab&space;$$,&space;which&space;is&space;constant.&space;Therefore,&space;$$&space;(a&space;-&space;b)^2&space;$$,&space;$$&space;(a^2&space;+&space;b^2)&space;$$,&space;and&space;$$&space;(a&space;+&space;b)^2&space;$$&space;are&space;in&space;arithmetic&space;progression.$

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Answered on 16 Apr Learn Arithmetic Progression

To find the common difference (d) of an arithmetic progression (AP), you can use the formula: read more

To find the common difference (d) of an arithmetic progression (AP), you can use the formula:
$T$&space;d&space;=&space;a_{n+1}&space;-&space;a_n&space;$&space;where&space;$$a_{n+1}$$&space;is&space;the&space;$$(n+1)$$th&space;term&space;and&space;$$a_n$$&space;is&space;the&space;$$n$$th&space;term&space;of&space;the&space;AP.&space;Alternatively,&space;if&space;you&space;know&space;the&space;first&space;term&space;($$a_1$$),&space;the&space;$$n$$th&space;term&space;($$a_n$$),&space;and&space;the&space;total&space;number&space;of&space;terms&space;($$n$$),&space;you&space;can&space;use&space;the&space;formula:&space;$&space;d&space;=&space;\frac{{a_n&space;-&space;a_1}}{{n&space;-&space;1}}&space;$&space;Let's&space;say&space;you&space;have&space;an&space;AP&space;given&space;by&space;the&space;terms&space;$$a_1,&space;a_2,&space;a_3,&space;\ldots,&space;a_n$$,&space;where&space;$$a_1$$&space;is&space;the&space;first&space;term&space;and&space;$$a_n$$&space;is&space;the&space;$$n$$th&space;term.&space;You&space;can&space;find&space;the&space;common&space;difference&space;using&space;either&space;of&space;these&space;formulas.$

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Answered on 16 Apr Learn Arithmetic Progression

To find out how many terms of the arithmetic progression 45,39,33,…45,39,33,… must be taken so that their sum is 180180, we can use the formula for the sum of the first nn terms of an arithmetic progression: read more

To find out how many terms of the arithmetic progression 45,39,33,…45,39,33,… must be taken so that their sum is 180180, we can use the formula for the sum of the first nn terms of an arithmetic progression:
$$&space;S_n&space;=&space;\frac{n}{2}&space;\times&space;(a_1&space;+&space;a_n)&space;$&space;where:&space;-&space;$$&space;S_n&space;$$&space;is&space;the&space;sum&space;of&space;the&space;first&space;$$&space;n&space;$$&space;terms,&space;-&space;$$&space;a_1&space;$$&space;is&space;the&space;first&space;term,&space;-&space;$$&space;a_n&space;$$&space;is&space;the&space;$$&space;n&space;$$th&space;term,&space;and&space;-&space;$$&space;n&space;$$&space;is&space;the&space;number&space;of&space;terms.&space;Given&space;that&space;$$&space;a_1&space;=&space;45&space;$$&space;and&space;the&space;common&space;difference&space;$$&space;d&space;=&space;39&space;-&space;45&space;=&space;-6&space;$$,&space;we&space;can&space;find&space;$$&space;a_n&space;$$&space;by&space;substituting&space;$$&space;n&space;$$&space;into&space;the&space;formula:&space;$&space;a_n&space;=&space;a_1&space;+&space;(n&space;-&space;1)&space;\times&space;d&space;$&space;Since&space;the&space;sum&space;of&space;the&space;terms&space;is&space;$$180$$,&space;we&space;can&space;set&space;up&space;the&space;equation:&space;$&space;180&space;=&space;\frac{n}{2}&space;\times&space;(a_1&space;+&space;a_n)&space;$&space;Substituting&space;the&space;values&space;of&space;$$&space;a_1&space;$$,&space;$$&space;a_n&space;$$,&space;and&space;$$&space;S_n&space;$$&space;into&space;the&space;equation,&space;we&space;get:&space;$&space;180&space;=&space;\frac{n}{2}&space;\times&space;\left(45&space;+&space;\left(45&space;+&space;(n&space;-&space;1)&space;\times&space;(-6)\right)\right)&space;$&space;Now,&space;let's&space;solve&space;for&space;$$&space;n&space;$$:&space;$&space;180&space;=&space;\frac{n}{2}&space;\times&space;(45&space;+&space;45&space;-&space;6n&space;+&space;6)&space;$&space;$&space;180&space;=&space;\frac{n}{2}&space;\times&space;(90&space;-&space;6n&space;+&space;6)&space;$&space;$&space;180&space;=&space;\frac{n}{2}&space;\times&space;(96&space;-&space;6n)&space;$&space;$&space;180&space;=&space;\frac{n}{2}&space;\times&space;(96&space;-&space;6n)&space;$&space;$&space;180&space;=&space;\frac{n}{2}&space;\times&space;(96&space;-&space;6n)&space;$&space;$&space;180&space;=&space;\frac{n}{2}&space;\times&space;(96&space;-&space;6n)&space;$&space;$&space;180&space;=&space;\frac{n}{2}&space;\times&space;(96&space;-&space;6n)&space;$&space;$&space;180&space;=&space;48n&space;-&space;3n^2&space;$&space;$&space;0&space;=&space;3n^2&space;-&space;48n&space;+&space;180&space;$&space;$&space;0&space;=&space;n^2&space;-&space;16n&space;+&space;60&space;$$

$Now,&space;we&space;can&space;use&space;the&space;quadratic&space;formula&space;to&space;find&space;the&space;values&space;of&space;$$&space;n&space;$$:&space;$&space;n&space;=&space;\frac{-b&space;\pm&space;\sqrt{b^2&space;-&space;4ac}}{2a}&space;$&space;where&space;$$&space;a&space;=&space;1&space;$$,&space;$$&space;b&space;=&space;-16&space;$$,&space;and&space;$$&space;c&space;=&space;60&space;$$.&space;$&space;n&space;=&space;\frac{-(-16)&space;\pm&space;\sqrt{(-16)^2&space;-&space;4&space;\times&space;1&space;\times&space;60}}{2&space;\times&space;1}&space;$&space;$&space;n&space;=&space;\frac{16&space;\pm&space;\sqrt{256&space;-&space;240}}{2}&space;$&space;$&space;n&space;=&space;\frac{16&space;\pm&space;\sqrt{16}}{2}&space;$&space;$&space;n&space;=&space;\frac{16&space;\pm&space;4}{2}&space;$&space;$&space;n_1&space;=&space;\frac{16&space;+&space;4}{2}&space;=&space;\frac{20}{2}&space;=&space;10&space;$&space;$&space;n_2&space;=&space;\frac{16&space;-&space;4}{2}&space;=&space;\frac{12}{2}&space;=&space;6&space;$&space;So,&space;there&space;are&space;two&space;possible&space;values&space;of&space;$$&space;n&space;$$:&space;$$&space;10&space;$$&space;and&space;$$&space;6&space;$$.&space;The&space;reason&space;for&space;the&space;double&space;answer&space;is&space;that&space;the&space;quadratic&space;equation&space;$$&space;n^2&space;-&space;16n&space;+&space;60&space;=&space;0&space;$$&space;has&space;two&space;real&space;roots.&space;This&space;is&space;because&space;the&space;arithmetic&space;progression&space;could&space;be&space;extended&space;both&space;forward&space;and&space;backward&space;from&space;the&space;starting&space;term&space;$$45$$&space;to&space;reach&space;a&space;sum&space;of&space;$$180$$.&space;Therefore,&space;both&space;$$6$$&space;and&space;$$10$$&space;terms&space;can&space;be$

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Answered on 16 Apr Learn Arithmetic Progression

In an arithmetic progression read more

In an arithmetic progression $I(AP),&space;the&space;$$n$$th&space;term&space;($$a_n$$)&space;is&space;given&space;by&space;the&space;formula:&space;$&space;a_n&space;=&space;a_1&space;+&space;(n&space;-&space;1)&space;\cdot&space;d&space;$&space;where:&space;-&space;$$&space;a_n&space;$$&space;is&space;the&space;$$n$$th&space;term,&space;-&space;$$&space;a_1&space;$$&space;is&space;the&space;first&space;term,&space;-&space;$$&space;d&space;$$&space;is&space;the&space;common&space;difference,&space;and&space;-&space;$$&space;n&space;$$&space;is&space;the&space;term&space;number.&space;Given&space;that&space;$$&space;d&space;=&space;-4&space;$$&space;and&space;$$&space;a_7&space;=&space;4&space;$$,&space;we&space;can&space;find&space;$$&space;a_1&space;$$&space;using&space;the&space;formula&space;for&space;the&space;$$n$$th&space;term:&space;$&space;a_7&space;=&space;a_1&space;+&space;(7&space;-&space;1)&space;\cdot&space;(-4)&space;$&space;Substitute&space;the&space;given&space;values:&space;$&space;4&space;=&space;a_1&space;+&space;6&space;\cdot&space;(-4)&space;$&space;$&space;4&space;=&space;a_1&space;-&space;24&space;$&space;Now,&space;solve&space;for&space;$$&space;a_1&space;$$:&space;$&space;a_1&space;=&space;4&space;+&space;24&space;$&space;$&space;a_1&space;=&space;28&space;$&space;So,&space;the&space;first&space;term&space;of&space;the&space;arithmetic&space;progression&space;is&space;$$28$$.$

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Answered on 16 Apr Learn Arithmetic Progression

Let's denote the four consecutive numbers in the arithmetic progression read more

Let's denote the four consecutive numbers in the arithmetic progression$(AP)&space;as&space;$$a$$,&space;$$a&space;+&space;d$$,&space;$$a&space;+&space;2d$$,&space;and&space;$$a&space;+&space;3d$$,&space;where&space;$$a$$&space;is&space;the&space;first&space;term&space;and&space;$$d$$&space;is&space;the&space;common&space;difference.&space;The&space;sum&space;of&space;these&space;four&space;consecutive&space;numbers&space;is&space;given&space;as&space;$$32$$,&space;so&space;we&space;can&space;write:&space;$&space;a&space;+&space;(a&space;+&space;d)&space;+&space;(a&space;+&space;2d)&space;+&space;(a&space;+&space;3d)&space;=&space;32&space;$&space;Now,&space;let's&space;simplify&space;this&space;equation:&space;$&space;4a&space;+&space;6d&space;=&space;32&space;$&space;$&space;2a&space;+&space;3d&space;=&space;16&space;$$

$Now,&space;we're&space;given&space;that&space;the&space;ratio&space;of&space;the&space;product&space;of&space;the&space;first&space;and&space;the&space;last&space;term&space;to&space;the&space;product&space;of&space;two&space;middle&space;terms&space;is&space;$$7&space;:&space;15$$.&space;Mathematically,&space;we&space;can&space;represent&space;this&space;as:&space;$&space;\frac{(a)(a&space;+&space;3d)}{(a&space;+&space;d)(a&space;+&space;2d)}&space;=&space;\frac{7}{15}&space;$&space;$&space;\frac{a(a&space;+&space;3d)}{(a&space;+&space;d)(a&space;+&space;2d)}&space;=&space;\frac{7}{15}&space;$&space;Now,&space;let's&space;simplify&space;this&space;equation:&space;$&space;\frac{a^2&space;+&space;3ad}{a^2&space;+&space;3ad&space;+&space;2ad&space;+&space;2d^2}&space;=&space;\frac{7}{15}&space;$&space;$&space;\frac{a^2&space;+&space;3ad}{a^2&space;+&space;5ad&space;+&space;2d^2}&space;=&space;\frac{7}{15}&space;$&space;Now,&space;let's&space;cross&space;multiply:&space;$&space;15(a^2&space;+&space;3ad)&space;=&space;7(a^2&space;+&space;5ad&space;+&space;2d^2)&space;$&space;$&space;15a^2&space;+&space;45ad&space;=&space;7a^2&space;+&space;35ad&space;+&space;14d^2&space;$&space;$&space;15a^2&space;-&space;7a^2&space;+&space;45ad&space;-&space;35ad&space;=&space;14d^2&space;$&space;$&space;8a^2&space;+&space;10ad&space;=&space;14d^2&space;$&space;$&space;4a^2&space;+&space;5ad&space;=&space;7d^2&space;$$

$We&space;can&space;also&space;rewrite&space;the&space;equation&space;$$2a&space;+&space;3d&space;=&space;16$$&space;as&space;$$a&space;=&space;16&space;-&space;3d$$&space;and&space;substitute&space;this&space;expression&space;for&space;$$a$$&space;in&space;the&space;equation&space;$$4a^2&space;+&space;5ad&space;=&space;7d^2$$:&space;$&space;4(16&space;-&space;3d)^2&space;+&space;5(16&space;-&space;3d)d&space;=&space;7d^2&space;$&space;Now,&space;let's&space;solve&space;this&space;quadratic&space;equation&space;for&space;$$d$$.&space;After&space;finding&space;$$d$$,&space;we&space;can&space;find&space;$$a$$&space;using&space;$$a&space;=&space;16&space;-&space;3d$$.$

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Answered on 16 Apr Learn Arithmetic Progression

The difference between any two terms in an arithmetic progression (AP) is constant and equal to the common difference (d). read more

The difference between any two terms in an arithmetic progression (AP) is constant and equal to the common difference (d).$T&space;Given&space;that&space;$$a_{21}&space;-&space;a_7&space;=&space;84$$,&space;where&space;$$a_{21}$$&space;is&space;the&space;21st&space;term&space;and&space;$$a_7$$&space;is&space;the&space;7th&space;term,&space;we&space;can&space;write:&space;$a_{21}&space;-&space;a_7&space;=&space;(a_1&space;+&space;20d)&space;-&space;(a_1&space;+&space;6d)&space;=&space;84$&space;$a_1&space;+&space;20d&space;-&space;(a_1&space;+&space;6d)&space;=&space;84$&space;$20d&space;-&space;6d&space;=&space;84$&space;$14d&space;=&space;84$&space;Now,&space;let's&space;solve&space;for&space;$$d$$:&space;$d&space;=&space;\frac{84}{14}$&space;$d&space;=&space;6$&space;So,&space;the&space;common&space;difference&space;of&space;the&space;arithmetic&space;progression&space;is&space;$$6$$.$

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Answered on 16 Apr Learn Arithmetic Progression

In this arithmetic progression, the common difference is read more

In this arithmetic progression, the common difference is $$$&space;-\frac{3}{4}&space;$$&space;because&space;each&space;term&space;decreases&space;by&space;$$&space;\frac{3}{4}&space;$$.&space;We&space;want&space;to&space;find&space;the&space;first&space;term&space;that&space;becomes&space;negative.&space;Let's&space;denote&space;the&space;$$n$$th&space;term&space;of&space;the&space;progression&space;as&space;$$&space;a_n&space;$$.&space;The&space;$$n$$th&space;term&space;is&space;given&space;by:&space;$&space;a_n&space;=&space;a_1&space;+&space;(n&space;-&space;1)&space;\cdot&space;d&space;$&space;where:&space;-&space;$$&space;a_n&space;$$&space;is&space;the&space;$$n$$th&space;term,&space;-&space;$$&space;a_1&space;$$&space;is&space;the&space;first&space;term,&space;-&space;$$&space;d&space;$$&space;is&space;the&space;common&space;difference,&space;and&space;-&space;$$&space;n&space;$$&space;is&space;the&space;term&space;number.&space;Given&space;that&space;$$&space;a_1&space;=&space;20&space;$$&space;and&space;$$&space;d&space;=&space;-\frac{3}{4}&space;$$,&space;we&space;can&space;find&space;$$&space;n&space;$$&space;such&space;that&space;$$&space;a_n&space;<&space;0&space;$$.&space;$&space;a_n&space;=&space;20&space;+&space;(n&space;-&space;1)&space;\cdot&space;\left(-\frac{3}{4}\right)&space;<&space;0&space;$&space;$&space;20&space;-&space;\frac{3}{4}(n&space;-&space;1)&space;<&space;0&space;$&space;$&space;20&space;-&space;\frac{3n}{4}&space;+&space;\frac{3}{4}&space;<&space;0&space;$$

$Now,&space;let's&space;solve&space;this&space;inequality&space;for&space;$$&space;n&space;$$:&space;$&space;-\frac{3n}{4}&space;+&space;\frac{83}{4}&space;<&space;0&space;$&space;$&space;-3n&space;+&space;83&space;<&space;0&space;$&space;$&space;-3n&space;<&space;-83&space;$&space;$&space;3n&space;>&space;83&space;$&space;$&space;n&space;>&space;\frac{83}{3}&space;$&space;Since&space;$$&space;n&space;$$&space;must&space;be&space;a&space;positive&space;integer,&space;the&space;smallest&space;integer&space;greater&space;than&space;$$&space;\frac{83}{3}&space;$$&space;is&space;$$&space;28&space;$$.&space;So,&space;the&space;first&space;negative&space;term&space;appears&space;at&space;the&space;$$&space;28&space;$$th&space;term&space;of&space;the&space;progression.$

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Answered on 16 Apr Learn Arithmetic Progression

Let's denote the sum of the first nn terms of the first arithmetic progression read more

Let's denote the sum of the first nn terms of the first arithmetic progression $as&space;$$&space;S_1&space;$$&space;and&space;the&space;sum&space;of&space;the&space;first&space;$$&space;n&space;$$&space;terms&space;of&space;the&space;second&space;arithmetic&space;progression&space;(AP)&space;as&space;$$&space;S_2&space;$$.&space;We're&space;given&space;that&space;the&space;ratio&space;of&space;$$&space;S_1&space;$$&space;to&space;$$&space;S_2&space;$$&space;is&space;$$&space;(7n&space;+&space;1)&space;:&space;(4n&space;+&space;27)&space;$$.&space;Therefore,&space;we&space;can&space;write:&space;$&space;\frac{S_1}{S_2}&space;=&space;\frac{7n&space;+&space;1}{4n&space;+&space;27}&space;$&space;The&space;sum&space;of&space;the&space;first&space;$$&space;n&space;$$&space;terms&space;of&space;an&space;arithmetic&space;progression&space;can&space;be&space;expressed&space;as:&space;$&space;S&space;=&space;\frac{n}{2}(a_1&space;+&space;a_n)&space;$&space;where&space;$$&space;a_1&space;$$&space;is&space;the&space;first&space;term&space;and&space;$$&space;a_n&space;$$&space;is&space;the&space;$$&space;n&space;$$th&space;term.$

$Now,&space;let's&space;denote&space;the&space;$$&space;9&space;$$th&space;term&space;of&space;the&space;first&space;arithmetic&space;progression&space;as&space;$$&space;a_{1,9}&space;$$&space;and&space;the&space;$$&space;9&space;$$th&space;term&space;of&space;the&space;second&space;arithmetic&space;progression&space;as&space;$$&space;a_{2,9}&space;$$.&space;We&space;know&space;that&space;$$&space;S_1&space;=&space;\frac{9}{2}(a_1&space;+&space;a_{1,9})&space;$$&space;and&space;$$&space;S_2&space;=&space;\frac{9}{2}(a_2&space;+&space;a_{2,9})&space;$$.&space;Therefore,&space;we&space;can&space;rewrite&space;the&space;given&space;ratio&space;as:&space;$&space;\frac{\frac{9}{2}(a_1&space;+&space;a_{1,9})}{\frac{9}{2}(a_2&space;+&space;a_{2,9})}&space;=&space;\frac{7n&space;+&space;1}{4n&space;+&space;27}&space;$&space;$&space;\frac{a_1&space;+&space;a_{1,9}}{a_2&space;+&space;a_{2,9}}&space;=&space;\frac{7n&space;+&space;1}{4n&space;+&space;27}&space;$&space;Given&space;that&space;both&space;progressions&space;are&space;arithmetic,&space;we&space;have&space;$$&space;a_{1,9}&space;=&space;a_1&space;+&space;8d_1&space;$$&space;and&space;$$&space;a_{2,9}&space;=&space;a_2&space;+&space;8d_2&space;$$,&space;where&space;$$&space;d_1&space;$$&space;and&space;$$&space;d_2&space;$$&space;are&space;the&space;common&space;differences&space;of&space;the&space;first&space;and&space;second&space;progressions,&space;respectively.$

$Now,&space;let's&space;find&space;the&space;ratio&space;of&space;the&space;$$&space;9&space;$$th&space;terms:&space;$&space;\frac{a_1&space;+&space;a_1&space;+&space;8d_1}{a_2&space;+&space;a_2&space;+&space;8d_2}&space;=&space;\frac{7n&space;+&space;1}{4n&space;+&space;27}&space;$&space;$&space;\frac{2a_1&space;+&space;8d_1}{2a_2&space;+&space;8d_2}&space;=&space;\frac{7n&space;+&space;1}{4n&space;+&space;27}&space;$&space;$&space;\frac{a_1&space;+&space;4d_1}{a_2&space;+&space;4d_2}&space;=&space;\frac{7n&space;+&space;1}{4n&space;+&space;27}&space;$&space;Given&space;that&space;$$&space;S_1&space;=&space;\frac{n}{2}(a_1&space;+&space;a_{1,9})&space;$$&space;and&space;$$&space;S_2&space;=&space;\frac{n}{2}(a_2&space;+&space;a_{2,9})&space;$$,&space;it&space;follows&space;that:&space;$&space;\frac{S_1}{S_2}&space;=&space;\frac{a_1&space;+&space;a_{1,9}}{a_2&space;+&space;a_{2,9}}&space;$&space;Thus,&space;we&space;can&space;write:&space;$&space;\frac{a_1&space;+&space;a_{1,9}}{a_2&space;+&space;a_{2,9}}&space;=&space;\frac{7n&space;+&space;1}{4n&space;+&space;27}&space;$&space;So,&space;the&space;ratio&space;of&space;the&space;$$&space;9&space;$$th&space;terms&space;is&space;$$&space;\frac{7n&space;+&space;1}{4n&space;+&space;27}&space;$$.$

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Answered on 16 Apr Learn Arithmetic Progression

Let's denote the first term of the arithmetic progression (AP) as a and the common difference as d read more

Let's denote the first term of the arithmetic progression (AP) as a and the common difference as d$Given&space;that&space;the&space;4th&space;term&space;of&space;the&space;AP&space;is&space;zero,&space;we&space;can&space;express&space;this&space;as&space;$$a&space;+&space;3d&space;=&space;0$$,&space;since&space;the&space;4th&space;term&space;is&space;$$a&space;+&space;3d$$.&space;This&space;gives&space;us&space;the&space;equation:&space;$a&space;+&space;3d&space;=&space;0$&space;Now,&space;let's&space;express&space;the&space;11th&space;term&space;and&space;the&space;25th&space;term&space;of&space;the&space;AP&space;in&space;terms&space;of&space;$$a$$&space;and&space;$$d$$:&space;-&space;The&space;11th&space;term&space;is&space;$$a&space;+&space;10d$$.&space;-&space;The&space;25th&space;term&space;is&space;$$a&space;+&space;24d$$.&space;We&space;want&space;to&space;prove&space;that&space;the&space;25th&space;term&space;is&space;three&space;times&space;the&space;11th&space;term,&space;so&space;we&space;have:&space;$3(a&space;+&space;10d)&space;=&space;a&space;+&space;24d$&space;Now,&space;let's&space;solve&space;this&space;equation:&space;$3a&space;+&space;30d&space;=&space;a&space;+&space;24d$&space;$3a&space;-&space;a&space;=&space;24d&space;-&space;30d$&space;$2a&space;=&space;-6d$&space;$a&space;=&space;-3d$$

$We&space;already&space;know&space;from&space;the&space;information&space;given&space;that&space;$$a&space;+&space;3d&space;=&space;0$$,&space;so&space;substituting&space;$$a&space;=&space;-3d$$&space;into&space;this&space;equation:&space;$-3d&space;+&space;3d&space;=&space;0$&space;$0&space;=&space;0$&space;This&space;equation&space;is&space;true,&space;confirming&space;that&space;our&space;assumption&space;for&space;$$a$$&space;and&space;$$d$$&space;is&space;valid.$

$Therefore,&space;since&space;$$a&space;=&space;-3d$$,&space;the&space;25th&space;term&space;is&space;indeed&space;three&space;times&space;the&space;11th&space;term,&space;satisfying&space;the&space;given&space;condition.$

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