1. For an A.P., let first term be $a$ and common difference be $d$.

   Then: $a_n = a + (n-1)d$ and $S_n = \frac{n}{2}\left[2a + (n-1)d\right]$

2. Given: $a_6 = 7$ So, $a + 5d = 7 \quad \text{(1)}$

3. Also given: $S_7 = 7$ Using sum formula, $\frac{7}{2}(2a + 6d) = 7$ $\frac{7}{2}\cdot 2(a+3d) = 7$ $7(a+3d) = 7$ $a + 3d = 1 \quad \text{(2)}$

4. Subtract (2) from (1): $(a+5d) - (a+3d) = 7 - 1$ $2d = 6$ $d = 3$

5. Put $d=3$ in (2): $a + 3(3) = 1$ $a + 9 = 1$ $a = -8$

6. Now use $S_n = 700$: $\frac{n}{2}[2(-8) + (n-1)3] = 700$ $\frac{n}{2}[-16 + 3n - 3] = 700$ $\frac{n}{2}(3n - 19) = 700$ $n(3n - 19) = 1400$ $3n^2 - 19n - 1400 = 0$

7. Solve the quadratic: $3n^2 - 19n - 1400 = 0$ Discriminant: $\Delta = (-19)^2 + 4\cdot 3\cdot 1400 = 361 + 16800 = 17161 = 131^2$

   So, $n = \frac{19 \pm 131}{6}$

   Positive value: $n = \frac{150}{6} = 25$

8. Therefore, $a_n = a_{25} = a + 24d$ $a_{25} = -8 + 24\cdot 3 = -8 + 72 = 64$

9. Checking options:

   • A: $65$ ❌
   • B: $56$ ❌
   • C: $70$ ❌
   • D: $64$ ✅

Hence, the required term is $64$.