1. We need all two-digit numbers which leave remainder $2$ or $5$ when divided by $7$.

2. Two-digit numbers are from $10$ to $99$.

A number leaving remainder $2$ on division by $7$ is of the form $n=7k+2$ and leaving remainder $5$ is of the form $n=7k+5$

Case 1: Numbers of the form $7k+2$

Smallest two-digit such number: $16=7\cdot 2+2$ Largest two-digit such number: $93=7\cdot 13+2$

So the sequence is: $16,23,30,37,44,51,58,65,72,79,86,93$ This is an AP with:

• first term $a=16$
• last term $l=93$
• common difference $d=7$

Number of terms: $n=\frac{93-16}{7}+1=\frac{77}{7}+1=11+1=12$

Sum: $S_1=\frac{n}{2}(a+l)=\frac{12}{2}(16+93)=6\cdot 109=654$

Case 2: Numbers of the form $7k+5$

Smallest two-digit such number: $12=7\cdot 1+5$ Largest two-digit such number: $96=7\cdot 13+5$

So the sequence is: $12,19,26,33,40,47,54,61,68,75,82,89,96$ This is an AP with:

• first term $a=12$
• last term $l=96$
• common difference $d=7$

Number of terms: $n=\frac{96-12}{7}+1=\frac{84}{7}+1=12+1=13$

Sum: $S_2=\frac{n}{2}(a+l)=\frac{13}{2}(12+96)=\frac{13}{2}\cdot 108=13\cdot 54=702$

Total sum

$S=S_1+S_2=654+702=1356$

4. Now compare with the options:

• A: $1356$ ✅
• B: $1256$ ❌
• C: $1365$ ❌
• D: $1465$ ❌

Therefore, the correct answer is: $\boxed{1356}$