1. Tardigrade
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  4. Let a , b , c (in that order) are in arithmetic progression and ( a /2), b , ( c /4) (in that order) are in geometric progression. If b=2, then the value of (a4+b4+c4) is equal to

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Q. Let $a , b , c$ (in that order) are in arithmetic progression and $\frac{ a }{2}, b , \frac{ c }{4}$ (in that order) are in geometric progression. If $b=2$, then the value of $\left(a^4+b^4+c^4\right)$ is equal to

Sequences and Series

Solution:

Given $a + c =2 b =4$ and $b ^2=\frac{ ac }{8} \Rightarrow ac =32$
Now
$a^4+c^4=\left(a^2+c^2\right)^2-2 a^2 c^2$
$=\left(( a + c )^2-2 ac \right)^2-2 a ^2 c ^2=(16-64)^2-2(32)^2=(48)^2-2(32)^2=(16)^2(9-8) $
$=256$
Hence $\left(a^4+b^4+c^4\right)=256+16=272$