Question

If the sum of the series $20+19 \frac{3}{5}+19 \frac{1}{5}+18 \frac{4}{5}+\ldots $ upto $n^ {th }$ term is $488$ and the $n ^ {th }$ term is negative, then :

• A. $n^ {th }$ term is $-4 \frac{2}{5}$
• B. $n= 41$
• C. $n^ {th }$ term is $-4$
• D. $n =60$

Answer 1

The Correct Option is C

The given problem involves determining the n^{th} term of an arithmetic progression (AP) from the series 20, 19 \frac{3}{5}, 19 \frac{1}{5}, 18 \frac{4}{5}, \ldots with its sum up to n^{th} term being 488 and the n^{th} term being negative. Let’s solve it step-by-step.

1. Identify the first term (a) and the common difference (d) of the series:
   • First term a = 20
   • Second term is 19 \frac{3}{5} = \frac{98}{5}
   The common difference can be calculated as:
   • d = \left(\frac{98}{5} - 20\right) = -\frac{2}{5}
2. Write the formula for the n^{th} term of an AP:
   • T_n = a + (n-1)d
3. Since the n^{th} term is negative, set the equation:
   • a + (n-1)d = -4
   • Substitute a = 20 and d = -\frac{2}{5}:
     • 20 + (n-1) \left(-\frac{2}{5}\right) = -4
     • 20 - \frac{2(n-1)}{5} = -4
     • Simplify and solve for n:
       • \frac{2(n-1)}{5} = 24
       • 2(n-1) = 120
       • 2n - 2 = 120
       • 2n = 122
       • n = 61
4. Check the sum formula for validation:
   • Sum of first n terms: S_n = \frac{n}{2} \left[2a + (n-1)d\right]
   • Set S_{61} = 488 and check:
     • \frac{61}{2} \left[2 \cdot 20 + 60 \left(-\frac{2}{5}\right)\right] = 488
     • \frac{61}{2} \left[40 - 24\right] = 488
     • \frac{61}{2} \cdot 16 = 488
     • 61 \cdot 8 = 488 matches!
5. However, since we also need to check if any other possible condition applies (like approximate integer part interpretation, aspect of problem miscalculations, specific series problem setup in tests):
   • Re-evaluate specific possible options / terms: If n = 41, ensure what the patterns show for consistency apart from calculation paths (noting possible setup error, checking question or option error as hypothetical checks).

Hence, confirming the process, the n^{th} term that is negative applies at the demonstrated potential mismatch marks, leading to the stated option: n^{th} term is -4.