Sequences & Series
AP and GP Conditions — 9(a²+b²+c²)
nta_pyq_2026_jan
Grade 11
Question:
Suppose $a,b,c$ are in A.P. and $a^2,2b^2,c^2$ are in G.P. If $a<b<c$ and $a+b+c=1$, then $9(a^2+b^2+c^2)$ is equal to _____.
Step-by-Step Solution
Key Concept: AP: $a+c=2b$. GP: $(2b^2)^2=a^2c^2\Rightarrow 2b^2=\pm ac$. Since $a<b<c$ (one negative possible): $2b^2=-ac$. Sum: $3b=1\Rightarrow b=1/3$. $a+c=2/3$, $ac=-2/9$.
$9(a^2+b^2+c^2)=9$.
Correct Answer: 9