1. **State the problem:** We have a geometric progression (GP) where the sum of the 2nd and 5th terms is 72, and the sum of the 3rd and 6th terms is 144. We need to find the common ratio $r$, the first term $a$, and the sum of the first 6 terms. 2. **Recall the formula for the $n$th term of a GP:** $$T_n = a r^{n-1}$$ 3. **Write the given conditions using the formula:** - Sum of 2nd and 5th terms: $$T_2 + T_5 = a r^{1} + a r^{4} = 72$$ - Sum of 3rd and 6th terms: $$T_3 + T_6 = a r^{2} + a r^{5} = 144$$ 4. **Factor out $a r$ and $a r^{2}$ respectively:** $$a r (1 + r^{3}) = 72$$ $$a r^{2} (1 + r^{3}) = 144$$ 5. **Divide the second equation by the first to eliminate $a$ and $(1 + r^{3})$:** $$\frac{a r^{2} (1 + r^{3})}{a r (1 + r^{3})} = \frac{144}{72}$$ $$\cancel{a} \frac{r^{2}}{\cancel{a} r} \cancel{(1 + r^{3})} / \cancel{(1 + r^{3})} = 2$$ $$r = 2$$ 6. **Substitute $r=2$ back into the first equation to find $a$:** $$a \cdot 2 (1 + 2^{3}) = 72$$ $$a \cdot 2 (1 + 8) = 72$$ $$a \cdot 2 \cdot 9 = 72$$ $$18a = 72$$ $$a = \frac{72}{18} = 4$$ 7. **Find the sum of the first 6 terms using the sum formula for GP:** $$S_n = a \frac{r^{n} - 1}{r - 1}$$ $$S_6 = 4 \frac{2^{6} - 1}{2 - 1} = 4 \frac{64 - 1}{1} = 4 \times 63 = 252$$ **Final answers:** - Common ratio $r = 2$ - First term $a = 4$ - Sum of first 6 terms $S_6 = 252$