The first- second and last terms of an A-P- are - - - respectively then the sum of first n terms is

The correct option is D ( α + γ)( β + γ 2α)/2( β α) (1) d=a_2 a_1= β α (2) a_n=a_1+(n 1)d γ =α +(n 1)(β α ) γ α/β α+1=n ...

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Standard X

Mathematics

Sum of N Terms of an AP

The first, se...

Question

The first, second and last terms of an A.P. are $α, β, γ$ respectively then the sum of first n terms is

β + γ - 2 α

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\frac{β + γ - 2 α}{β - α}

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\frac{β + γ + 2 α}{β + α}

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\frac{(α + γ) (β + γ - 2 α)}{2 (β - α)}

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Solution

The correct option is D $\frac{(α + γ) (β + γ - 2 α)}{2 (β - α)}$
(1) $d = a_{2} - a_{1} = β - α$
(2) $a_{n} = a_{1} + (n - 1) d$
$γ = α + (n - 1) (β - α)$
$\frac{γ - α}{β - α} + 1 = n$
$\frac{γ + β - 2 α}{β - α} = n$
so, $S_{n} = \frac{n}{α} [a_{1} + a_{n}]$
$= \frac{(γ + β - 2 α) (α + γ)}{2 (β - α)}$

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The first, second and last terms of an A.P. are α,β,γ respectively then the sum of first n terms is

The correct option is D (α+γ)(β+γ−2α)2(β−α)(1) d=a2−a1=β−α(2) an=a1+(n−1)dγ=α+(n−1)(β−α)γ−αβ−α+1=nγ+β−2αβ−α=nso, Sn=nα[a1+an]=(γ+β−2α)(α+γ)2(β−α)

The first, second and last terms of an A.P. are $α, β, γ$ respectively then the sum of first n terms is