tolong bantu pake cara yaa

a = bilangan pertama = [tex]a+ \frac{1}{2} \sqrt{2} [/tex]
b = beda bilangan = [tex](a+ \frac{3}{2} \sqrt{2})-(a+ \frac{1}{2} \sqrt{2} ) = \sqrt{2} [/tex]

Sn = [tex] \frac{n}{2} [/tex](2a + (n - a)b)
S₁₀ = [tex] \frac{10}{2} [/tex](2a + (10-1) × b)
[tex]55 \sqrt{2}=5(2( a+ \frac{1}{2} \sqrt{2} )+9. \sqrt{2}) [/tex]
55√2 ÷ 5 = 2(a + [tex] \frac{1}{2} \sqrt{2} [/tex]) + 9√2
11√2 = 2(a + [tex] \frac{1}{2} \sqrt{2} [/tex]) + 9√2
11√2 - 9√2 = 2(a + [tex] \frac{1}{2} \sqrt{2} [/tex])
2√2 = 2(a + [tex] \frac{1}{2} \sqrt{2} [/tex])
2√2 ÷ 2 = a + [tex] \frac{1}{2} \sqrt{2} [/tex]
√2 = a + [tex] \frac{1}{2} \sqrt{2} [/tex]
√2 = a + [tex] \frac{ \sqrt{2} }{2} [/tex]
a = [tex] \sqrt{2} - \frac{ \sqrt{2} }{2} = \frac{2 \sqrt{2} }{2} - \frac{ \sqrt{2} }{2} = \frac{1}{2} \sqrt{2} [/tex]
(C)