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The first term of an A.P is 14 and the sums of first five terms and first ten terms are equal is magnitude but opposite in sign. The 3rd term of the AP is?
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The first term of an A.P is 14 and the sums of first five terms and fi...
Understanding the Problem
The first term of an arithmetic progression (A.P) is given as 14. We know that the sums of the first five terms and the first ten terms are equal in magnitude but opposite in sign. Let’s denote the common difference as \(d\).

Formulating the Sums
The sum of the first \(n\) terms of an A.P can be calculated using the formula:
\[
S_n = \frac{n}{2} \times (2a + (n-1)d)
\]
Here, \(a\) is the first term and \(d\) is the common difference.

Calculating \(S_5\) and \(S_{10}\)
- **Sum of the first five terms \(S_5\)**:
\[
S_5 = \frac{5}{2} \times (2 \times 14 + 4d) = \frac{5}{2} \times (28 + 4d) = 10 + 10d
\]
- **Sum of the first ten terms \(S_{10}\)**:
\[
S_{10} = \frac{10}{2} \times (2 \times 14 + 9d) = 5 \times (28 + 9d) = 140 + 45d
\]

Setting Up the Equation
According to the problem, \(S_5\) and \(S_{10}\) are equal in magnitude but opposite in sign:
\[
10 + 10d = -(140 + 45d)
\]

Simplifying the Equation
Expanding and rearranging the equation:
\[
10 + 10d + 140 + 45d = 0
\]
\[
55d + 150 = 0
\]
\[
d = -\frac{150}{55} = -\frac{30}{11}
\]

Finding the 3rd Term
The 3rd term of the A.P is given by:
\[
T_3 = a + 2d = 14 + 2\left(-\frac{30}{11}\right) = 14 - \frac{60}{11} = \frac{154 - 60}{11} = \frac{94}{11}
\]

Conclusion
Thus, the 3rd term of the A.P is:
\[
\frac{94}{11} \approx 8.545
\]
This gives the required value of the third term in the arithmetic progression.
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The first term of an A.P is 14 and the sums of first five terms and first ten terms are equal is magnitude but opposite in sign. The 3rd term of the AP is?
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