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The first term of an AP is 14 and the sums of the first five terms and the 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 AP is 14 and the sums of the first five terms and...
Problem: Find the 3rd term of an AP given that the first term is 14 and the sums of the first five terms and the first ten terms are equal in magnitude but opposite in sign.

Solution:

Step 1: Find the common difference

Let the common difference of the AP be d.

The sum of the first five terms of the AP is given by:

S5 = (5/2)(2a + 4d)

where a = 14 (first term)

S5 = (5/2)(28 + 4d)

The sum of the first ten terms of the AP is given by:

S10 = (10/2)(2a + 9d)

S10 = (10/2)(28 + 9d)

Since the sums of the first five and ten terms are equal in magnitude but opposite in sign, we have:

S5 = -S10

(5/2)(28 + 4d) = -(10/2)(28 + 9d)

Simplifying the above equation, we get:

20d = -3(28)

d = -42/20 = -21/10

Therefore, the common difference of the AP is -21/10.

Step 2: Find the 3rd term

The 3rd term of the AP can be found using the formula:

a3 = a + 2d

Substituting a = 14 and d = -21/10, we get:

a3 = 14 + 2(-21/10)

a3 = 14 - 21/5

a3 = 49/5

Therefore, the 3rd term of the AP is 49/5.

Conclusion: The 3rd term of the AP is 49/5.
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The first term of an AP is 14 and the sums of the first five terms and...
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The first term of an AP is 14 and the sums of the first five terms and the first ten terms are equal is magnitude but opposite in sign. The 3rd term of the Ap is?
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