====== 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.
=====Question 17=====
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 $1224$.
=====Question 18=====
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 $162$.
=====Question 19=====
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 $-220$.GOOD
====Go to ====
[[math-11-nbf:sol:unit04:ex4-3-p8|< Question 15 & 16]]
[[math-11-nbf:sol:unit04:ex4-3-p10|Question 20, 21 & 22 >]]