Let $a_n$ be the $n$ th term of an A.P. If $S_n = a_1 + a_2 + \dots + a_n = 700$ for some $n$, $a_6 = 7$ and $S_7 = 7$, then $a_n$ is equal to:

Step 1: Formulate equations from given information

We are given an arithmetic progression (A.P.). The general formula for the $k$ -th term of an A.P. is $a_k = a + (k-1)d$, where $a$ is the first term and $d$ is the common difference. The sum of the first $k$ terms of an A.P. is given by $S_k = \frac{k}{2}(2a + (k-1)d)$. We will use these formulas to set up equations based on the given conditions.

Step 2: Solve for 'a' and 'd' using given conditions

We are given $a_6 = 7$ and $S_7 = 7$. Using the formulas from the previous step, we can write these as two linear equations in terms of $a$ and $d$. From $a_6 = 7$, we get $a + 5d = 7$. From $S_7 = 7$, we get $\frac{7}{2}(2a + (7-1)d) = 7$, which simplifies to $7(a + 3d) = 7$, and further to $a + 3d = 1$. Now we have a system of two linear equations.

Step 3: Find the common difference 'd' and first term 'a'

To find $d$, we subtract equation (2) from equation (1): $(a + 5d) - (a + 3d) = 7 - 1$, which simplifies to $2d = 6$, so $d = 3$. Now substitute $d=3$ into equation (2): $a + 3(3) = 1$, which gives $a + 9 = 1$, so $a = -8$.

Step 4: Find 'n' using the sum formula

We are given $S_n = 700$. Substitute the values of $a = -8$ and $d = 3$ into the sum formula: $S_n = \frac{n}{2}(2(-8) + (n-1)3) = 700$. This simplifies to a quadratic equation: $3n^2 - 19n - 1400 = 0$.

Step 5: Solve the quadratic equation for 'n'

We solve the quadratic equation $3n^2 - 19n - 1400 = 0$ using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Substituting the coefficients, we get $n = \frac{19 \pm \sqrt{(-19)^2 - 4(3)(-1400)}}{2(3)}$. This simplifies to $n = \frac{19 \pm \sqrt{361 + 16800}}{6} = \frac{19 \pm \sqrt{17161}}{6} = \frac{19 \pm 131}{6}$. Since $n$ must be a positive integer, we take the positive root: $n = \frac{19 + 131}{6} = \frac{150}{6} = 25$. The negative root is rejected.

Step 6: Calculate the 'n'th term, $a_n$

Now that we have $a = -8$, $d = 3$, and $n = 25$, we can find the $n$ -th term, $a_n$, using the formula $a_n = a + (n-1)d$. Substituting the values, we get $a_{25} = -8 + (25-1)3 = -8 + 24 \times 3 = -8 + 72 = 64$.

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