Find five number in Arithmetic progressions whose sum is 25 and the su

1.	Find five number in Arithmetic progressions whose sum is 25 and the sum of whose square is 135
Answer» SOLUTION : $\bf{\red{\underline{\underline{\bf{Given\::}}}}}$ Five NUMBER in ARITHMETIC progression WHOSE sum is 25 and the sum of whose square is 135. $\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$ The number. $\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$ Let the numbers be a-2d, a-d, a, a+d, a+2d A/q $\mapsto\sf{a-2d+a-d+a+a+d+a+2d=25}\\\\\mapsto\sf{5a \cancel{-3d+3d}=25}\\\\\mapsto\sf{5a=25}\\\\\mapsto\sf{a=\cancel{\dfrac{25}{5} }}\\\\\mapsto\sf{\pink{a=5}}$ So; Putting the value of (a) in this number : $\mapsto\sf{(a-2d)^{2} +(a-d)^{2} +a^{2} +(a+d)^{2} +(a+2d)^{2} =135}\\\\\mapsto\sf{a^{2} +4d^{2} \cancel{-4ad}+a^{2} +d^{2} \cancel{-2ad}+a^{2} +a^{2} +d^{2} \cancel{+2ad}+a^{2} +4d^{2} \cancel{+4ad}=135}\\\\\mapsto\sf{5a^{2} +10d^{2} =135}\\\\\mapsto\sf{5(5)^{2} +10d^{2}=135}\\ \\\mapsto\sf{5\times 25+10d^{2} =135}\\\\\mapsto\sf{125+10d^{2} =135}\\\\\mapsto\sf{10d^{2} =135-125}\\\\\mapsto\sf{10d^{2} =10}\\\\\mapsto\sf{d^{2} =\cancel{\dfrac{10}{10} }}\\\\\mapsto\sf{d^{2} =1}\\\\$ $\mapsto\sf{\pink{d=\pm1}}$ Now; Putting the value of d = 1 a-2d = 5 - 2(1) = 5 - 2 = 3 a-d = 5 - 1 = 4 a = 5 a+d = 5 + 1 = 6 a + 2d = 5 + 2(1) = 5+2 = 7 Putting the value of d = -1 a - 2d = 5 - 2(-1) = 5 + 2 = 7 a - d = 5 - (-1) = 5 + 1 = 6 a = 5 a + d = 5 + (-1) = 5 - 1 = 4 a + 2d = 5 + 2(-1) = 5 - 2 = 3

Find five number in Arithmetic progressions whose sum is 25 and the sum of whose square is 135

Answer»

SOLUTION :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

Five NUMBER in ARITHMETIC progression WHOSE sum is 25 and the sum of whose square is 135.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The number.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

Let the numbers be a-2d, a-d, a, a+d, a+2d

A/q

$\mapsto\sf{a-2d+a-d+a+a+d+a+2d=25}\\\\\mapsto\sf{5a \cancel{-3d+3d}=25}\\\\\mapsto\sf{5a=25}\\\\\mapsto\sf{a=\cancel{\dfrac{25}{5} }}\\\\\mapsto\sf{\pink{a=5}}$

So;

Putting the value of (a) in this number :

$\mapsto\sf{(a-2d)^{2} +(a-d)^{2} +a^{2} +(a+d)^{2} +(a+2d)^{2} =135}\\\\\mapsto\sf{a^{2} +4d^{2} \cancel{-4ad}+a^{2} +d^{2} \cancel{-2ad}+a^{2} +a^{2} +d^{2} \cancel{+2ad}+a^{2} +4d^{2} \cancel{+4ad}=135}\\\\\mapsto\sf{5a^{2} +10d^{2} =135}\\\\\mapsto\sf{5(5)^{2} +10d^{2}=135}\\ \\\mapsto\sf{5\times 25+10d^{2} =135}\\\\\mapsto\sf{125+10d^{2} =135}\\\\\mapsto\sf{10d^{2} =135-125}\\\\\mapsto\sf{10d^{2} =10}\\\\\mapsto\sf{d^{2} =\cancel{\dfrac{10}{10} }}\\\\\mapsto\sf{d^{2} =1}\\\\$

$\mapsto\sf{\pink{d=\pm1}}$

Now;

Putting the value of d = 1

a-2d = 5 - 2(1) = 5 - 2 = 3
a-d = 5 - 1 = 4
a = 5
a+d = 5 + 1 = 6
a + 2d = 5 + 2(1) = 5+2 = 7

Putting the value of d = -1

a - 2d = 5 - 2(-1) = 5 + 2 = 7
a - d = 5 - (-1) = 5 + 1 = 6
a = 5
a + d = 5 + (-1) = 5 - 1 = 4
a + 2d = 5 + 2(-1) = 5 - 2 = 3