Question 17, 18 and 19, Exercise 4.3

Solutions of Question 17, 18 and 19 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find sum of the arithmetic series. $6+12+18+\ldots+96$.

Solution.

Given arithmetic series: $$6+12+18+\ldots+96.$$ So, $a_{1}=6$, $d=12-6=6$, $a_{n}=96$, $n=?$.
We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & 96=6+(n-1)(6) \\ \implies & 96=6+6n-6 \\ \implies & 6n=96 \\ \implies & n = 24. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{24}&=\frac{24}{2}[6+96]\\ &=12\times 102\\ &=1224. \end{align} Hence the sum of given series is $1224$.

Find sum of the arithmetic series. $34+30+26+\ldots+2$

Solution. Given arithmetic series: $$34+30+26+\ldots+2.$$ So, $a_{1}=34$, $d=30-34=-4$, $a_{n}=2$, $n=?$. We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & 2=34+(n-1)(-4) \\ \implies & 2=34-4n+4 \\ \implies & 2=38-4n \\ \implies & 4n=36 \\ \implies & n = 9. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{9}&=\frac{9}{2}[34+2]\\ &=\frac{9}{2}\times 36\\ &=162. \end{align} Hence the sum of the given series is $162$.

Find sum of the arithmetic series. $10+4+(-2)+\ldots+(-50)$

Solution.

Given arithmetic series: $$10+4+(-2)+\ldots+(-50).$$ So, $a_{1}=10$, $d=4-10=-6$, $a_{n}=-50$, $n=?$. We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & -50=10+(n-1)(-6) \\ \implies & -50=10-6n+6 \\ \implies & 6n=16+50 \\ \implies & 6n= 66 \\ \implies & n = 11. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{11}&=\frac{11}{2}[10+(-50)]\\ &=\frac{11}{2}\times (-40)\\ &=-220. \end{align} Hence the sum of the given series is $-220$.

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