Question

In an A.P., if the 5^th term is \( \frac{1}{7} \) and the 7^th term is \( \frac{1}{5} \), then the sum of the first 35 terms is ______. 

• A. $9$
• B. $18$
• C. $36$
• D. $72$
• E. $83$

Hint:

When two non-consecutive terms of an A.P. are given, subtract their equations to quickly eliminate $a$ and find $d$. Then substitute back to find $a$.

Answer 1

The Correct Option is B

Concept:
An Arithmetic Progression (A.P.) is a sequence in which each term differs from the previous one by a constant difference $d$. Key formulas used:
• $n^{\text{th}}$ term: $a_n = a + (n-1)d$
• Sum of first $n$ terms: $S_n = \frac{n}{2}\left[2a + (n-1)d\right]$ Here, $a$ is the first term and $d$ is the common difference.

Step 1: Write expressions for the given terms.
The general A.P. is: \[ a, \; a+d, \; a+2d, \; a+3d, \; a+4d, \; a+5d, \; a+6d, \ldots \] Thus, \[ a_5 = a + 4d = \frac{1}{7} \cdots (1) \] \[ a_7 = a + 6d = \frac{1}{5} \cdots (2) \]

Step 2: Eliminate $a$ to find the common difference $d$.
Subtract equation (1) from (2): \[ (a+6d) - (a+4d) = \frac{1}{5} - \frac{1}{7} \] Simplify: \[ 2d = \frac{7 - 5}{35} = \frac{2}{35} \] \[ d = \frac{1}{35} \]

Step 3: Substitute $d$ back to find the first term $a$.
Using equation (1): \[ a + 4d = \frac{1}{7} \] \[ a + 4\left(\frac{1}{35}\right) = \frac{1}{7} \] Convert $\frac{1}{7}$ into denominator 35: \[ \frac{1}{7} = \frac{5}{35} \] So, \[ a + \frac{4}{35} = \frac{5}{35} \] \[ a = \frac{5}{35} - \frac{4}{35} = \frac{1}{35} \]

Step 4: Write the A.P. explicitly (for clarity).
\[ \frac{1}{35}, \; \frac{2}{35}, \; \frac{3}{35}, \; \frac{4}{35}, \; \frac{5}{35}, \ldots \]

Step 5: Use sum formula to compute $S_{35$.}
\[ S_{35} = \frac{35}{2}\left[2a + (35-1)d\right] \] Substitute values: \[ S_{35} = \frac{35}{2}\left[2\left(\frac{1}{35}\right) + 34\left(\frac{1}{35}\right)\right] \] \[ = \frac{35}{2} \left(\frac{2 + 34}{35}\right) \] \[ = \frac{35}{2} \cdot \frac{36}{35} \] Cancel 35: \[ = \frac{36}{2} = 18 \]
Final Answer: \[ \boxed{18} \]