If a_{n} = 3 - 4n,show that a_{1}, a_{2}, a_{ | vedclass

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(B) Given that the $n^{th}$ term of the series is $a_{n} = 3 - 4n$.For $n = 1$,$a_{1} = 3 - 4(1) = 3 - 4 = -1$.For $n = 2$,$a_{2} = 3 - 4(2) = 3 - 8 =

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$-780$

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(B) Given that the $n^{th}$ term of the series is $a_{n} = 3 - 4n$.For $n = 1$,$a_{1} = 3 - 4(1) = 3 - 4 = -1$.For $n = 2$,$a_{2} = 3 - 4(2) = 3 - 8 = -5$.For $n = 3$,$a_{3} = 3 - 4(3) = 3 - 12 = -9$.For $n = 4$,$a_{4} = 3 - 4(4) = 3 - 16 = -13$.So,the series is $-1, -5, -9, -13, \ldots$We observe that the common difference $d$ is:$a_{2} - a_{1} = -5 - (-1) = -4$$a_{3} - a_{2} = -9 - (-5) = -4$$a_{4} - a_{3} = -13 - (-9) = -4$Since the difference between consecutive terms is constant $(d = -4)$,the series forms an $AP$.The sum of $n$ terms of an $AP$ is given by $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.For $n = 20$,$a = -1$,and $d = -4$:$S_{20} = \frac{20}{2}[2(-1) + (20 - 1)(-4)]$$S_{20} = 10[-2 + (19)(-4)]$$S_{20} = 10[-2 - 76]$$S_{20} = 10 	imes (-78) = -780$.Thus,the sum of $20$ terms is $-780$.

If $a_{n} = 3 - 4n$,show that $a_{1}, a_{2}, a_{3}, \ldots$ form an $AP$. Also,find $S_{20}$.

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