Question

find the three numbers in AP such that their sum is 24 and the sum of their square is 194

Answer 1

let three numbers in ap are x, x+1, x+2,
A/Q x+x+1+ x+2 =24
3x+3=24
x=7
after putting the value
the numbers are 7 , 8 ,and 9

hope it helps you

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

${\boxed{\sf\:{First\;term\;be\;p}}}$

Also

${\boxed{\sf\:{Common\;difference\;be\;d}}}$

Hence,

Sum of three number of AP

p - d + p + p + d = 24

3p = 24

$\tt{\rightarrow p=\dfrac{24}{3}}$

p = 8

Sum of their squares is 194

Hence

(p - d)² + (p)² + (p + d)² = 194

$\large{\boxed{\sf\:{Substitute\;the\;value\;of\;p}}}$

(8 - d)² + (8)² + (8 + d)² = 194

64 - 2 × 8 × d + d² + 64 + 64 + 2 × 8 × d + d² = 194

64 - 16d + d² + 64 + 64 + 16d + d² = 194

2d² + 192 = 194

2d² = 194 - 192

2d² = 2

$\tt{\rightarrow d^2=\dfrac{2}{2}}$

d² = 1

d = ±1

Hence we get :-

${\boxed{\sf\:{For\;d=1}}}$

p - d = 8 - 1 = 7

p = 8

p + d = 8 + 1 = 9

${\boxed{\sf\:{For\;d=-1}}}$

p - d = 8 - (-1) = 9

p = 8

p + d = 8 + (-1) = 7

$\Large{\boxed{\sf\:{7,8,9\;or\;9,8,7}}}$