Let $$a_n$$ be the $$n^{\text{th}}$$ term of an A.P. If $$S_n = a_1 + a_2 + a_3 + \ldots + a_n = 700$$, $$a_6 = 7$$ and $$S_7 = 7$$, then $$a_{n}$$ is equal to :

For an arithmetic progression (A.P.) let the first term be $$a$$ and the common difference be $$d$$. Then the $$n^{\text{th}}$$ term is $$a_n = a + (n-1)d$$ and the sum of the first $$n$$ terms is $$S_n = \dfrac{n}{2}\,\bigl[2a + (n-1)d\bigr]$$.

We are given three pieces of information:

• $$a_6 = 7$$ • $$S_7 = 7$$     • $$S_n = 700$$ for some positive integer $$n$$.

Step 1: Use $$a_6 = 7$$.
$$a_6 = a + 5d = 7$$ $$-(1)$$

Step 2: Use $$S_7 = 7$$.
$$S_7 = \dfrac{7}{2}\,\bigl[2a + 6d\bigr] = 7$$ Divide both sides by $$7$$: $$\dfrac{1}{2}\,\bigl[2a + 6d\bigr] = 1 \;\Longrightarrow\; 2a + 6d = 2$$ Simplify: $$a + 3d = 1$$ $$-(2)$$

Step 3: Solve equations $$(1)$$ and $$(2)$$ for $$a$$ and $$d$$.
Subtract $$(2)$$ from $$(1)$$: $$(a + 5d) - (a + 3d) = 7 - 1 \;\Longrightarrow\; 2d = 6 \;\Longrightarrow\; d = 3$$ Substitute $$d = 3$$ into $$(1)$$: $$a + 5(3) = 7 \;\Longrightarrow\; a + 15 = 7 \;\Longrightarrow\; a = -8$$

Step 4: Find $$n$$ such that $$S_n = 700$$.
$$S_n = \dfrac{n}{2}\,\bigl[2a + (n-1)d\bigr]$$ Insert $$a = -8$$ and $$d = 3$$: $$S_n = \dfrac{n}{2}\,\bigl[2(-8) + (n-1)3\bigr] = \dfrac{n}{2}\,\bigl[-16 + 3n - 3\bigr] = \dfrac{n}{2}\,(3n - 19)$$ Set this equal to $$700$$: $$\dfrac{n}{2}\,(3n - 19) = 700 \;\Longrightarrow\; n(3n - 19) = 1400$$ This is a quadratic: $$3n^2 - 19n - 1400 = 0$$

Compute the discriminant:
$$\Delta = (-19)^2 - 4(3)(-1400) = 361 + 16800 = 17161$$ Since $$17161 = 131^2$$, the roots are $$n = \dfrac{19 \pm 131}{2 \times 3}$$ Taking the positive root: $$n = \dfrac{150}{6} = 25$$ (Discard the negative root as $$n$$ must be positive.)

Step 5: Find $$a_{25}$$.
$$a_{25} = a + 24d = -8 + 24 \times 3 = -8 + 72 = 64$$

Therefore, $$a_{25} = 64$$.
This matches Option C.