Question

Find four terms of the A.P. whose sum is -4 and the sum of whose squares

is 84​

Answer 1

Given :-

• Sum of four terms of AP is -4 and the sum of their squares are 84

To Find :-

• we have to find the 4 terms of AP

Solution

• Let the four terms of AP be

$\sf\pink{ a-3d,\ a-d,\ a+d,\ a+3d}$

$\underline{\bigstar{\sf\ According\ to\ Question :- }}$

$:\implies\sf\ a-\cancel{3d}+a-\cancel{d}+a+\cancel{d}+a+\cancel{3d}=-4\\ \\ \\ :\implies\sf\ 4a=-4\\ \\ \\ :\implies\underline{\boxed{\red{\sf\ a= -1}}}$

Now ,

$:\implies\sf\ (a-3d)^2+(a-d)^2+(a+d)^2+(a+3d)^2=84\\ \\ \\ :\implies\sf\ \big\{(a-3d)^2+(a+3d)^2\big\}+\big\{(a-d)^2+(a+d)^2\big\}=84\\ \\ \\ :\implies\sf\big\{a^2+9d^2-\cancel{6ad}+a^2+9d^2+\cancel{6ad}\big\}+\big\{a^2+d^2-\cancel{2ad}+a^2+d^2+\cancel{2ad}\big\}=84\\ \\ \\ :\implies\sf\ \big\{2a^2+18d^2+2a^2+2d^2\big\}=84\\ \\ \\ :\implies\sf\ 4(-1)^2+20d^2=84\ \ \ \ \ \ \therefore\ \Big[a= -1\Big]\\ \\ \\ :\implies\sf\ 4+20d^2=84\\ \\ \\ :\implies\sf\ 20d^2=84-4\\ \\ \\ :\implies\sf\ d^2= \cancel{\dfrac{80}{20}}\\ \\ \\ :\implies\sf d^2= 4\\ \\ \\ :\implies\sf\ d= \sqrt{4}\\ \\ \\ :\implies\boxed{\purple{\sf\ d=2}}$

$\bullet\sf\ a-3d= -1-3(2)= -1-6= (-7)\\ \\ \\\bullet\sf\ a-d= -1-2= -3\\ \\ \\ \bullet\sf\ a+d= -1+2= 1\\ \\ \\\bullet\sf\ a+3d= -1+3(2)= -1+6= 5$

$\underline{\boxed{\bigstar{\sf\ Terms\ of\ AP= -7,\ -3,\ 1,\ 5}}}$