The sum of the squares of two consecutive multiples of 7 is 1225

Let the required consecutive multiples of 7 be 7x and 7(x + 1).

According to the given condition,

$(7 x)^{2}+[7(x+1)]^{2}=1225$

$\Rightarrow 49 x^{2}+49\left(x^{2}+2 x+1\right)=1225$

$\Rightarrow 49 x^{2}+49 x^{2}+98 x+49=1225$

$\Rightarrow 98 x^{2}+98 x-1176=0$

$\Rightarrow x^{2}+x-12=0$

$\Rightarrow x^{2}+4 x-3 x-12=0$

$\Rightarrow x(x+4)-3(x+4)=0$

$\Rightarrow(x+4)(x-3)=0$

$\Rightarrow x+4=0$ or $x-3=0$

$\Rightarrow x=-4$ or $x=3$

∴ x = 3            (Neglecting the negative value)

When x = 3,
7x = 7 × 3 = 21
7(x + 1) = 7(3 + 1) = 7 × 4 = 28

Hence, the required multiples are 21 and 28.