Question 5 & 6 Exercise 4.3

Solutions of Question 5 & 6 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find four numbers in an arithmetic sequence, whose sum is $20$ and the sum of whose squares is $120$ .

Let the four numbers are
$$a-2 d, a-d, a+d, a+2 d,$$ $Condition-1$
Their sum is $20$ , thus
\begin{align}a-3 d+a-d+a+d+a+3 d&=20 \\ \Rightarrow 4 a&=20\\ \Rightarrow a&=5 .\end{align} $Condition-2$
The sum of their square is $120$, therefore
\begin{align}(a-3 d)^2+(a-d)^2+(a+d)^2+(a+2 d)^2&=120 \\ \Rightarrow a^2-6 a d+9 d^2+a^2-2 a d+d^2+a^2+2 a d+d^2+a^2+6 a d+9 d^2&=120 \\ \Rightarrow 4 a^2+20 d^2&=120 \\ \Rightarrow 20 d^2&=120-4 \cdot(5)^2 \because a=5 \\ \Rightarrow 20 d^2&=20 \\ \Rightarrow d^2&=1 \text { or } d= \pm 1\end{align} When $a=5$ and $d=1$ then the numbers are
\begin{align} a-3d&=5-3=2, \\ a-d&=5-1=4, \\ a+d&=5+1=6 \text { and } \\ a+3 d&=5+3=8.\end{align} When $a=5$ and $d=-1$ then the numbers are
\begin{align}a-3 d&=5-3(-1)=8, \\ a-d&=5-(-1)=6, \\ a+d&=5+(-1)=, 4 \text { and } \\ a+3 d&=5-3=2\\ & 2,4,6,8; 8,6,4,2\end{align}

If $x_1, x_2, x_3, \ldots$, are in A.P. If
\begin{align}x_1+x_7+x_{10}&=-6 \text { and } \\ x_3+x_8+x_{12}&=-11, \text { find } \\ x_3+x_8+x_{22} .\end{align}

We are given
\begin{align}x_1, x_2, x_3, \ldots, \text { an A.P } \\ x_1+x_7+x_{10}&=-6\end{align}
When $a=5$ and $d=-1$ then the numbers are
\begin{align}\therefore x_1+(x_1+6 d)+(x_1+9 d)&=-6 \\ \Rightarrow 3 x_1+15 d&=-6 \ldots \ldots \ldots \ldots . . .(i) \\ \text { Also } x_3+x_8+x_{12}&=-11 \\ \Rightarrow(x_1+2 d)+(x_1+7 d)+(x_1+11 d)& =-11 \\ \Rightarrow 3 x_1+20 d&=-11 \ldots \ldots \ldots \ldots . .(ii) \end{align} Subtracting (ii) from (i), we get
\begin{align}(3 x_1+15 d)-(3 x_1+20 d)&=-6-(-11) \\ \Rightarrow-5 d&=5 \\ \Rightarrow d&=-1.\end{align} Putting $d=-1$ in (i), we have
\begin{align}3 x_1+15(-1)&=-6 \\ \Rightarrow 3 x_1&=-6+15=9 \\ \Rightarrow \quad x_1&=\dfrac{9}{3}=3.\end{align} Now $$x_3=x_1+2 d=3+2(-1)=1$$ \begin{align}x_8&=x_1+7 d=3+7(-1)=-4, \text { and } \\ x_{22}&=x_1+21 d=3+21(-1)=-18 \\ \therefore \quad x_3+x_8+x_{22}&=1-4-18=-21\end{align}

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