Question

In an Arithmetic Progression Whose first term is 2,the sum of first five terms is one fourth the sum of the next five terms . Show ThatT20=-112 Find S20

Answer 1

Answer:

- 1100

Step-by-step explanation:

Given, a = 2    ;  

⇒ S₅ = (1/4) [T₆+ T₇ + T₈ + T₉ + T₁₀]

⇒ S₅ = (1/4) [S₁₀ - S₅]    ⇒ 4S₅ = S₁₀ - S₅

⇒ 5S₅ = S₁₀

⇒ 5 * (5/2)(2a + 4d) = (10/2) (2a + 9d)

⇒ 25(2a + 4d) = 10(2a + 9d)

⇒ 50a + 100d = 20a + 90d

⇒ -30a = 10d

⇒ -3a = d      ⇒ -3(2) = d        [a = 2]

⇒ -6 = d

   ∴ T₂₀ = a + 19d

             = 2 + 19(-6)

             = - 112             proved

    ∴ S₂₀ = (20/2) [2  + (-112)]

              = -1100

*read thoroughly ;   used:

  Sₙ = (n/2) [2a + (n - 1)d]

       or  (n/2) [a + l]  , l is last term

   Tₙ = a + (n - 1)d