1. **State the problem:** We have a frequency distribution with class intervals and frequencies: 0-20 (6), 20-40 (8), 40-60 ($f_1$), 60-80 (12), 80-100 (6), 100-120 (5). The mode is given as 65, which lies in the modal class 60-80 with frequency 12. We need to find $f_1$ given that the frequencies 6, 8, $f_1$, and 12 are in ascending order. 2. **Recall the mode formula for grouped data:** $$\text{Mode} = L + \frac{(f_m - f_{m-1})}{(2f_m - f_{m-1} - f_{m+1})} \times h$$ where: - $L$ = lower boundary of modal class - $f_m$ = frequency of modal class - $f_{m-1}$ = frequency of class before modal class - $f_{m+1}$ = frequency of class after modal class - $h$ = class width 3. **Identify values:** - Modal class: 60-80, so $L = 60$ - $f_m = 12$ - $f_{m-1} = f_1$ (frequency of 40-60 class) - $f_{m+1} = 6$ (frequency of 80-100 class) - $h = 20$ 4. **Plug in the mode value 65:** $$65 = 60 + \frac{(12 - f_1)}{(2 \times 12 - f_1 - 6)} \times 20$$ 5. **Simplify the equation:** $$65 - 60 = \frac{12 - f_1}{24 - f_1 - 6} \times 20$$ $$5 = \frac{12 - f_1}{18 - f_1} \times 20$$ 6. **Divide both sides by 20:** $$\frac{5}{20} = \frac{12 - f_1}{18 - f_1}$$ $$\frac{1}{4} = \frac{12 - f_1}{18 - f_1}$$ 7. **Cross multiply:** $$1 \times (18 - f_1) = 4 \times (12 - f_1)$$ $$18 - f_1 = 48 - 4f_1$$ 8. **Solve for $f_1$:** $$18 - f_1 = 48 - 4f_1$$ $$-f_1 + 4f_1 = 48 - 18$$ $$3f_1 = 30$$ $$f_1 = 10$$ 9. **Check ascending order condition:** Given frequencies 6, 8, $f_1$, 12 must be ascending. Check: 6 < 8 < 10 < 12 is true. **Final answer:** $$\boxed{10}$$