Question

Answer this question (◍•ᴗ•◍)❤​

Answer 1

Let the common difference of the AP's be d and let the first term he a-2d

now, AP will be

(a-2d),(a-d),(a),(a+d),(a+2d)

now, according to question

a-2d+a-d+a+a+d+a+2d=25

5a=25

a=5

required AP will be

(5-2d),(5-d),(5),(5+d),(5+2d)

now According to second condition

(5-2d)^2+(5-d)^2+25+(5+d)^2+(5+2d)^2=135

25+4d^2-20d+25+d^2-10d+25+25+d^2+10d+25+4d^2+20d=135

125+10d^2=135

d^2=1

d=+1 or -1

d=+1 or -1

This two value of d, shows that there will be 2 AP in the same required condition

1st AP=3,4,5,6,7,,

second AP=7,6,5,4,3

Answer 2

$\sf\blue{Question}$

$\sf{A \ straight \ wooden \ stick \ of \ length \ 25 \ cm}$

$\sf{is \ cut \ into \ 5 \ parts \ whose \ lengths \ are}$

$\sf{in \ AP. \ The \ sum \ of \ the \ squares \ of \ each}$

$\sf{part \ is \ 135. \ Find \ the \ length \ of \ each}$

$\sf{part \ of \ the \ stick.}$

___________________________________

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

$\sf{The \ length \ of \ part \ of \ each \ piece \ is}$

$\sf{3 \ cm, \ 4 \ cm, \ 5 \ cm, \ 6 \ cm \ and \ 7 \ cm}$

$\sf{respectively.}$

$\sf\orange{Given:}$

$\sf{\implies{Stick \ of \ 25 \ cm \ is \ cutted \ in \ 5 \ pieces}}$

$\sf{\implies{Length \ of \ the \ pieces \ are \ in \ AP}}$

$\sf{\implies{Sum \ of \ squares \ of \ each \ piece \ is \ 135}}$

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

$\sf{Length \ of \ each \ part \ of \ piece.}$

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

$\sf{Let \ length \ of \ pieces \ in \ AP \ be}$

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

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

$\sf{(a-2d)+(a-d)+a+(a+d)+(a+2d)=25}$

$\sf{5a=25}$

$\sf{a=\frac{25}{5}}$

$\sf{a=5...(1)}$

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

$\sf{(a-2d)^{2}+(a-d)^{2}+a^{2}+(a+d)^{2}+(a+2d)^{2}=135}$

$\sf{5a^{2}+10d^{2}=135}$

$\sf{But \ a=5 \ from \ (1)}$

$\sf{5(5)^{2}+10d^{2}=135}$

$\sf{5\times25+10d^{2}=135}$

$\sf{125+10d^{2}=135}$

$\sf{10d^{2}=135-125}$

$\sf{\therefore{10d^{2}=10}}$

$\sf{\therefore{d^{2}=\frac{10}{10}}}$

$\sf{d^{2}=1}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{d=1}$

$\sf{Length \ of \ Pieces:}$

$\sf{a-2d=5-2(1)=3 \ cm,}$

$\sf{a-d=5-1=4 \ cm,}$

$\sf{a=5 \ cm,}$

$\sf{a+d=5+1=6 \ cm,}$

$\sf{a+2d=7 \ cm.}$

$\sf\purple{\tt{\therefore{The \ length \ of \ part \ of \ each \ piece \ is}}}$

$\sf\purple{\tt{3 \ cm, \ 4 \ cm, \ 5 \ cm, \ 6 \ cm \ and \ 7 \ cm}}$

$\sf\purple{\tt{respectively.}}$