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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 =

Vedclass · Accepted Answer

$-780$

Vedclass · Answer

(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 $S_{20}$.

Similar Questions

How many terms of the $A.P.$ $5, 8, 11, \dots$ add up to $670$?

The $n^{th}$ term of an $A.P.$ is given by $T_{n} = 5 - 6n$. Find the sum of the first $n$ terms of the $A.P.$

Find the sum of the first $15$ multiples of $8.$

For an $A.P.$,the first term is $10$ and the $10^{th}$ term is $100$. Then,the sum of the first $10$ terms is..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module