Answer :

mhanifa

12

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Let the first term of the arithmetic sequence be a and the common difference be d.

The nth term of the arithmetic sequence can be expressed as

  • aₙ = a + (n - 1)d

The 1st, 5th, and 10th terms of this arithmetic sequence are:

  • a₁ = a,
  • a₅ = a + 4d,
  • a₁₀ = a + 9d

Since these terms form a GP, the ratio between consecutive terms is constant. Therefore, we get the equation:

  • (a + 4d)/a = (a + 9d)/(a + 4d)
  • (a + 4d)² = a*(a + 9d)
  • a² + 8ad + 16d² = a² + 9ad
  • 8ad + 16d² = 9ad
  • 16d² = ad

Since d is not zero, we can divide both sides by d:

  • 16d = a

We are also given that the sum of the 2nd and 8th terms is 30. Therefore:

  • a₂ = a + d
  • a₈ = a + 7d

So:

  • (a + d) + (a + 7d) = 30

Substituting a = 16d gives:

  • (16d + d) + (16d + 7d) = 30
  • 40d = 30

Solving for d, we get:

  • d = 30/40 = 3/4

Now, substituting back to find a:

  • a = 16d = 16*(3/4) = 12

The first term a of the arithmetic sequence is 12.

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