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  • If an = 3 - 4n, show that a1, a2, a3 ,... form an AP. Also find S20.

If  a_n = 3 - 4n, \text{show that} a\textsubscript{1}, a\textsubscript{2}, a\textsubscript{3} ,... form an AP. Also find S\textsubscript{20}

Answers (1)

Here it is given that

 a\textsubscript{n} = 3 - 4n\\

Put n =1 then

 a\textsubscript{1} = 3 - 4(1)\\

a\textsubscript{1} = 3 - 4= - 1\\

Put n =2 then

 a\textsubscript{2} = 3 - 4(2)\\

a\textsubscript{2} = 3 \textbf{-} 8 = \textbf{- }5\\  

Put n =3 then

 a\textsubscript{3} = 3 - 4(3)\\

a\textsubscript{3} = 3 - 12 = - 9\\  

\text{Then A.P is -1, -5, -9} \ldots \ldots \\ a\textsubscript{2} - a\textsubscript{1}= - 5 - (-1) = - 5 + 1 = - 4\\ a\textsubscript{3} - a\textsubscript{2} = - 9 - (-5) = - 9 + 5 = - 4\\ a\textsubscript{2} - a\textsubscript{1} = a\textsubscript{3} - a\textsubscript{2}\\ \text{ Hence}, a\textsubscript{1}, a\textsubscript{2}, a\textsubscript{3} \ldots \text{ forms an AP with } a\textsubscript{1 }= a = - 1 , d = -4\\ \text{ Here d= common difference and first term is a} = a\textsubscript{1}= - 1\\S_{2}=\frac{20}{2}\left[ 2\left( -1 \right)+\left( 20-1 \right)\left( -4 \right) \right] \left( Using \ S_{n}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \right) \\ = 10 [- 2 + 19(-4)]\\ = 10[-2 - 76]\\ S\textsubscript{20} = - 780\\

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