Question

pls pls pls pls pls pls pls pls pls pls pls anzwer Q-The sum of three numbers in A.P. is 36 and the sum of their squares is 450. Find the numbers.
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Answer 1

$\sf\blue{Question:}$

$\sf{The \ sum \ of \ three \ numbers \ in \ A.P.}$

$\sf{is \ 36 \ and \ the \ sum \ of \ their \ squares}$

$\sf{is \ 450. \ Find \ the \ numbers.}$

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$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ numbers \ are \ 9, \ 12 \ and \ 15 \ or}$

$\sf{15, \ 12 \ and \ 9.}$

$\sf\orange{Given:}$

$\sf{\implies{The \ sum \ of \ three \ numbers \ in \ A.P.}}$

$\sf{is \ 36 \ and}$

$\sf{\implies{Sum \ of \ their \ squares \ is \ 450}}$

$\sf\pink{To \ find:}$

$\sf{The \ numbers.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ the \ three \ numbers \ of \ A.P. \ be}$

$\sf{(a-d), \ a \ and \ (a+d)}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{(a-d)+a+(a+d)=36}$

$\sf{\therefore{3a=36}}$

$\sf{a=\frac{36}{3}}$

$\sf{\therefore{a=12...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{(a-d)^{2}+a^{2}+(a+d)^{2}}$

$\sf{By \ identity}$

$\sf{(a-b)^{2}=a^{2}-2ab+b^{2}}$

$\sf{(a+b)^{2}=a^{2}+2ab+b^{2}}$

$\sf{3a^{2}+d^{2}=450}$

$\sf{But \ a=12 \ from \ equation (1)}$

$\sf{3(12)^{2}+2d^{2}=450}$

$\sf{432+2d^{2}=450}$

$\sf{2d^{2}=450-432}$

$\sf{2d^{2}=18}$

$\sf{\therefore{d^{2}=\frac{18}{2}}}$

$\sf{\therefore{d^{2}=9}}$

$\sf{On \ taking \ square \ root \ on \ both \ sides.}$

$\sf{d=3 \ or \ -3}$

$\sf{If \ d =3 \ numbers \ are:}$

$\sf{a-d=12-3=9}$

$\sf{a=12}$

$\sf{a+d=12+3=16}$

$\sf{Numbers \ are \ 9, \ 12 \ and \ 15.}$

$\sf{If \ d=-3 \ numbers \ are:}$

$\sf{a-d=12-(-3)=15}$

$\sf{a=12}$

$\sf{a+d=12+(-3)=9}$

$\sf{Numbers \ are \ 14, \ 11 \ and \ 8.}$

$\sf\purple{\tt{\therefore{The \ numbers \ are \ 9, \ 12 \ and \ 15 \ or}}}$

$\sf\purple{\tt{15, \ 12 \ and \ 9.}}$

Answer 2

$\color{darkblue}\underline{\underline{\sf Question:}}$

$\rm{Find \ three \ numbers \ in \ A.P. \ such}$

$\rm{that \ their \ sum \ is \ 36 \ and \ the \ sum \ of \ their}$

$\rm{squares \ is \ 450}$

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$\color{green}\underline{\underline{\sf Answer:}}$

$\sf\large\underline\purple{The \ three \ numbers \ are \ 9 , \ 12 ,\ 15}$

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$\color{red}\underline{\underline{\sf given:}}$

$\rm{The \ sum \ of \ three \ numbers : \ 36}$

$\rm{\ Sum \ of \ square \ of \ three \ numbers : \ 450}$

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$\color{orange}\underline{\underline{\sf To \ find:}}$

$\rm{Find \ the \ three \ numbers}$

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$\color{blue}\underline{\underline{\sf solution:}}$

$\sf\large\underline{let ,}$$\rm/{The \ three \ numbers \ are :}$

$a \: (a + d) \: (a + 2d)$

$\rm{As \ pr \ given,}$

$a + (a + d) + (a + 2d) = 36$

........................................(1)

$\rm{and,}$

${a}^{2} + {(a + d)}^{2} + {(a + 2d)}^{2} = 450$

........................................(2)

$\rm{According \ to \ first \ condition}$

$3a + 3d = 36$

∴$3(a + d) = 36$

∴$a + d = \frac{36}{3}$

∴$a + d = 12$

$\rm{According \ to \ second \ condition}$

${a}^{2} + {12}^{2} + {(12 + d)}^{2} = 450$

∴${a}^{2} + 144 + 144 + 24d + {d}^{2} = 450$

∴${a}^{2} + 24d + {d}^{2} = 450 - 144 - 144$

∴${a}^{2} + 24d + {d}^{2} = 162$

$\rm{We \ know ,}$

$d = 12 - a$

$\rm{putting \ this \ value \ in \ above \ equation}$

${(12 - d)}^{2} + 24d + {d}^{2} = 162$

⇒$144 - 24d + 24d + {d}^{2} + {d}^{2} = 162$

⇒$2( {d}^{2} )= 162 - 144 = 18$

${d}^{2} = \frac{18}{2}$

⇒${d}^{2} = 9$

⇒$d = 3$

⇒$a = 12 - 3 = 9$

$\rm{The \ three \ numbers \ are,}$

$a = 9$

⇒$a + d = 12$

⇒$a + 2d = 9 + 2(3) = 15$

$\sf\large\underline\purple{The \ three \ numbers \ are \ 9 , \ 12 ,\ 15}$

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$\rm\red{Hope \ it \ helps \ :))}$