If the first four terms of an arithmetic sequence are: a, 2a, b and a-6-b for some
numbers “a” and “b”, then the value of th

1 thought on “If the first four terms of an arithmetic sequence are: a, 2a, b and a-6-b for some<br /> numbers “a” and “b”, then the value of th”

1. Answer:

   The value of the 100 term is – 100.

   Step-by-step explanation:

   Here we know that a, 2a, b and a-6-b are in arithmetic progression.

   First find out the common difference,

   • d = a₂ – a₁
   • d = 2a – a
   • d = a

   • d = a₃ – a₂
   • d = b – 2a

   Substitute the value of d,

   • a = b – 2a
   • b = a + 2a
   • b = 3a

   Now,

   • a₄ = a + 3d

   We know, d = a

   • a – 6 – b = a + 3a
   • a – 6 – b = a + 3a
   • a – 6 – b = 4a

   Substituting the value of b,

   • a – 6 – 3a = 4a
   • – 6 = 4a + 3a – a
   • – 6 = 4a + 2a
   • – 6 = 6a
   • a = -6/6
   • a = – 1

   Now, we know that d = a which is also equal to – 1.

   • a₁₀₀ = a + 99d

   Again d = a

   • a₁₀₀ = a + 99a
   • a₁₀₀ = 100a

   Substituting the value of a,

   • a₁₀₀ = 100 × – 1
   • a₁₀₀ = – 100

   ∴ The value of the 100 term is – 100.

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