1. **State the problem:** We are given an arithmetic progression (A.P.) with terms $T_7 = 23$, $T_n = 43$, and $T_{2n} = 91$. We need to find the first term $a$, the common difference $d$, and the term number $n$. 2. **Recall the formula for the $k$-th term of an A.P.:** $$T_k = a + (k-1)d$$ where $a$ is the first term and $d$ is the common difference. 3. **Write equations using the given terms:** - For $T_7 = 23$: $$a + 6d = 23$$ - For $T_n = 43$: $$a + (n-1)d = 43$$ - For $T_{2n} = 91$: $$a + (2n-1)d = 91$$ 4. **Use the first equation to express $a$ in terms of $d$:** $$a = 23 - 6d$$ 5. **Substitute $a$ into the second and third equations:** - Second equation: $$23 - 6d + (n-1)d = 43$$ Simplify: $$23 - 6d + nd - d = 43$$ $$23 + (n - 7)d = 43$$ $$ (n - 7)d = 20$$ - Third equation: $$23 - 6d + (2n - 1)d = 91$$ Simplify: $$23 - 6d + 2nd - d = 91$$ $$23 + (2n - 7)d = 91$$ $$ (2n - 7)d = 68$$ 6. **From the two equations involving $d$ and $n$:** $$ (n - 7)d = 20$$ $$ (2n - 7)d = 68$$ Divide the second equation by the first to eliminate $d$: $$\frac{(2n - 7)d}{(n - 7)d} = \frac{68}{20}$$ $$\frac{2n - 7}{n - 7} = 3.4$$ 7. **Solve for $n$:** $$2n - 7 = 3.4(n - 7)$$ $$2n - 7 = 3.4n - 23.8$$ $$-7 + 23.8 = 3.4n - 2n$$ $$16.8 = 1.4n$$ $$n = \frac{16.8}{1.4} = 12$$ 8. **Find $d$ using $ (n - 7)d = 20$:** $$ (12 - 7)d = 20$$ $$5d = 20$$ $$d = 4$$ 9. **Find $a$ using $a = 23 - 6d$:** $$a = 23 - 6 \times 4 = 23 - 24 = -1$$ **Final answers:** $$a = -1, \quad d = 4, \quad n = 12$$