GATE 2026 DA Q.62: solution and answer

Official answer: 10Source: Official answer key, GATE 2026 DA (IIT Guwahati)

1. Let $S_{xx}={\sum }_{i=1}^n(x_i-\bar{x}{)}^2$. The double sum is a pairwise squared-difference sum, so rewrite it through the mean.

2. Insert $\bar{x}$: $(x_i-x_j{)}^2=((x_i-\bar{x})-(x_j-\bar{x}){)}^2=(x_i-\bar{x}{)}^2-2(x_i-\bar{x})(x_j-\bar{x})+(x_j-\bar{x}{)}^2$.

3. Sum over $i$ and $j$. The first term gives $n{S}_{xx}$, because it does not depend on $j$ and the sum over $j$ contributes a factor $n$. The last term gives $n{S}_{xx}$ in the same way. The cross term gives $-2\left({\sum }_i(x_i-\bar{x})\right)\left({\sum }_j(x_j-\bar{x})\right)=0$, since deviations from the mean sum to zero.

4. Hence $$\sum _{i=1}^n\sum _{j=1}^n(x_i-x_j{)}^2=2n{S}_{xx}.$$

5. The given condition says this sum equals $2000\times 99=198000$. With $n=100$: $200S_{xx}=198000$, so $S_{xx}=990$.

6. Required value: $\frac{1}{99}{S}_{xx}=\frac{990}{99}=10$.