Question:

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

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When solving problems involving A.P., use the formula for the sum and the general term to find unknowns, and simplify the system of equations to find the solution.

Updated On: Oct 31, 2025

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The Correct Option is C

Approach Solution - 1

To solve the problem, let's analyze the information given and make use of the formulas related to an arithmetic progression (A.P.). The problem states the following:

The sum of the first $n$ terms of the A.P., denoted as $S_n$, is 700 for some $n$.
The 6th term, $a_6$, is 7.
The sum of the first 7 terms, $S_7$, is 7.

The sum of the first $n$ terms of an A.P. is given by:

$S_n = \frac{n}{2} \left(2a + (n-1)d\right)$

where $a$ is the first term, $d$ is the common difference, and $n$ is the number of terms.

Let's first use the condition: $S_7 = 7$. Plugging $n = 7$ into the formula for the sum, we have:

$S_7 = \frac{7}{2} (2a + 6d) = 7$

Simplifying gives:

$7a + 21d = 14 \quad \Rightarrow \quad 7a + 21d = 14$

Dividing by 7, we get:

$a + 3d = 2 \quad \Rightarrow \quad (1)$

Now, using the condition: $a_6 = 7$.

The nth term of an A.P. is given by:

$a_n = a + (n-1)d$

For $a_6$, we have:

$a + 5d = 7 \quad \Rightarrow \quad (2)$

We now have two equations:

(1) $a + 3d = 2$
(2) $a + 5d = 7$

Subtract equation (1) from equation (2):

$(a + 5d) - (a + 3d) = 7 - 2$

$2d = 5$

Solve for $d$:

$d = \frac{5}{2}$

Substitute $d = \frac{5}{2}$ back into equation (1):

$a + 3 \times \frac{5}{2} = 2$

$a + \frac{15}{2} = 2$

$a = 2 - \frac{15}{2}$

$a = -\frac{11}{2}$

Now, using the sum condition $\displaystyle S_n = 700$ in the equation:

$700 = \frac{n}{2} \left(2(-\frac{11}{2}) + (n-1)\frac{5}{2}\right)$

Simplifying, we get:

$700 = \frac{n}{2}\left(-11 + \frac{5n-5}{2}\right)$

$700 = \frac{n}{2}\left(\frac{5n-27}{2}\right)$

$700 \times 4 = n(5n - 27)$

$2800 = 5n^2 - 27n$

$5n^2 - 27n - 2800 = 0$

Solve this quadratic equation for $n$ using the quadratic formula.

After finding the feasible $n$, we can calculate $a_n$:

$a_n = a + (n-1)d$

Substitute the appropriate values of $a$ and $d$ to find:

$a_n = 64$

Thus, the correct answer is 64.

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Approach Solution -2

Step 1: General Form of an A.P.
An arithmetic progression (A.P.) has the general form: \[ a_k = a_1 + (k-1)d \] where:

$ a_1 $ is the first term,
$ d $ is the common difference.

The sum of the first $ n $ terms ($ S_n $) is given by: \[ S_n = \frac{n}{2} [2a_1 + (n-1)d] \] \subsection{Step 2: Use Given Information $ a_6 = 7 $} Given the 6th term: \[ a_6 = a_1 + 5d = 7 \quad \text{(1)} \]
Step 3: Use Given Information $ S_7 = 7 $
Given the sum of the first 7 terms: \[ S_7 = \frac{7}{2} [2a_1 + 6d] = 7 \] Simplify: \begin{align} \frac{7}{2} [2a_1 + 6d] &= 7 [2a_1 + 6d] &= 2 \quad \text{(Divide both sides by 7/2)} a_1 + 3d &= 1 \quad \text{(Divide by 2)} \quad \text{(2)} \end{align}
Step 4: Solve for $ a_1 $ and $ d $
From equation (2): \[ a_1 = 1 - 3d \quad \text{(3)} \] Substitute (3) into equation (1): \begin{align} (1 - 3d) + 5d &= 7 1 + 2d &= 7 \\2d &= 6 \\d &= 3 \end{align} Now substitute $ d = 3 $ back into equation (3): \[ a_1 = 1 - 3(3) = -8 \]
Step 5: Find $ n $ for $ S_n = 700 $
Using the sum formula with $ a_1 = -8 $ and $ d = 3 $: \[ S_n = \frac{n}{2} [2(-8) + (n-1)(3)] = 700 \] Simplify: \begin{align} \frac{n}{2} [-16 + 3n - 3] &= 700 \frac{n}{2} [3n - 19] &= 700 n(3n - 19) &= 1400 \\3n^2 - 19n - 1400 &= 0 \end{align} Solve the quadratic equation: \[ n = \frac{19 \pm \sqrt{(-19)^2 - 4 \cdot 3 \cdot (-1400)}}{2 \cdot 3} \] \[ n = \frac{19 \pm \sqrt{361 + 16800}}{6} \] \[ n = \frac{19 \pm \sqrt{17161}}{6} \] \[ n = \frac{19 \pm 131}{6} \] Possible solutions:

$ n = \frac{19 + 131}{6} = 25 $
$ n = \frac{19 - 131}{6} $ (negative, discard)

Thus, $ n = 25 $.
Step 6: Find $ a_n $ (the $ n $-th term)
Using the general form: \[ a_n = a_1 + (n-1)d \] For $ n = 25 $: \begin{align} a_{25} &= -8 + (25-1) \times 3 &= -8 + 72 \\&= 64 \end{align}
Conclusion
The value of $ a_n $ is $ 64 $, which corresponds to option 3.