find the sum of all 21 terms of an AP who consist 21 terms the sum of the three terms in the middle is 129 and of the last three terms is 237.
Answers
Answer:
903
Step-by-step explanation:
Given :
Sum of the three middle terms = 129
Sum of the last three terms = 237
Procedure :
The middle term of the AP = (n + 1)/2 th term
Hence here the middle term is 11th term.
Hence a₁₀ + a₁₁ + a₁₂ = 129
As aₓ = a + (x - 1)d
⇒ a + 9d + a + 10d + a + 11d = 129
⇒ 3a + 30d = 129
∴ a + 10d = 43
The last three terms of the AP are a₁₉,a₂₀ and a₂₁.
Hence a₁₉ + a₂₀ + a₂₁ = 237
⇒ a + 18d + a + 19d + a + 20d = 237
⇒ 3a + 57d = 237
∴ a + 19d = 79
Hence, from both the equations :
(a + 19d) - (a + 10d) = 79 - 43
⇒ 9d = 36
∴ d = 4
Substitute d value in the first equation.
a + 10d = 43
⇒ a + 10(4) = 43
⇒ a + 40 = 43
∴ a = 3
Hence the first term of the AP is 3, and the common difference is 4.
As sₓ = (x/2)[2a + (x - 1)d]
⇒ s₂₁ = (21/2)[2(3) + (21 - 1)(4)]
⇒ s₂₁ = (21/2)[6 + 80]
⇒ s₂₁ = (21/2)(86)
⇒ s₂₁ = 21 × 43
⇒ s₂₁ = 903
Hence the sum of all 21 terms of the AP = 903.
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