1. **Problem Statement:** Given an arithmetic progression (A.P.) with terms $T_7 = 23$, $T_n = 43$, and $T_{2n} = 91$, find the first term $a$, the common difference $d$, and the term number $n$. 2. **Formula for the $k$-th term of an A.P.:** $$T_k = a + (k-1)d$$ 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. **Subtract the first equation from the second:** $$(a + (n-1)d) - (a + 6d) = 43 - 23$$ $$ (n-1)d - 6d = 20$$ $$ (n-7)d = 20 \\ ext{(Equation 1)}$$ 5. **Subtract the second equation from the third:** $$(a + (2n-1)d) - (a + (n-1)d) = 91 - 43$$ $$ (2n-1)d - (n-1)d = 48$$ $$ (2n - 1 - n + 1)d = 48$$ $$ nd = 48 \\ ext{(Equation 2)}$$ 6. **From Equation 1:** $$ (n-7)d = 20$$ 7. **From Equation 2:** $$ nd = 48$$ 8. **Express $d$ from Equation 2:** $$ d = \frac{48}{n}$$ 9. **Substitute $d$ into Equation 1:** $$ (n-7) \times \frac{48}{n} = 20$$ $$ 48 \times \frac{n-7}{n} = 20$$ $$ 48(n-7) = 20n$$ $$ 48n - 336 = 20n$$ $$ 48n - 20n = 336$$ $$ 28n = 336$$ $$ n = \frac{336}{28} = 12$$ 10. **Find $d$ using $n=12$:** $$ d = \frac{48}{12} = 4$$ 11. **Find $a$ using $T_7 = 23$:** $$ a + 6d = 23$$ $$ a + 6 \times 4 = 23$$ $$ a + 24 = 23$$ $$ a = 23 - 24 = -1$$ **Final answers:** $$ a = -1, \quad d = 4, \quad n = 12$$