# Question 3 & 4 Exercise 4.3

Solutions of Question 3 & 4 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 sum of all the numbers divisible by $5$ from $25$ to $350$.

The numbers divisible by $5$ from $25$ tò $350$ are
$$25,30,35, \ldots, 350.$$ This is A.P. with $a_1=25, d=5$ and $a_n=350$
To find $n$, we know that \begin{align}a_n&=a_1+(n-1) d\end{align} in the given case it becomes,
\begin{align} 350&=25+(n-1)(5) \\ \Rightarrow 5 n-5+25&=350 \\ \Rightarrow 5 n&=350-20=330 \\ \Rightarrow n&=66, \text { now for the sum } \\ S_n&=\dfrac{n}{2}(a_1+a_n), \text { that becomes } \\ S_{66}&=\dfrac{66}{2}(25+350) \\ \Rightarrow S_{66}&=33(375)=12375 .\end{align}

The sum of three numbers in an arithmetic sequence is $36$ and the sum of their cubes is $6336$ . Find them.

Let us suppose the three numbers are $a-d, a, a+d$\\. then by first condition their sum is equal to $36$
\begin{align}(a-d)+a+(a+d)&=36 \\ \Rightarrow 3 a&=36 \\ \Rightarrow a&=12\end{align} Now by the second condition, the sum of their cubes is $6336$,
so we have
\begin{align}(a-d)^3+a^3+(a+d)^3&=6336 \\ \Rightarrow a^3-3 a^2 d+3 a d^2-d^3+a^3+a^3+3 a^2 d+3 a d^2+d^3&=6336 \\ \Rightarrow 3 a^3+6 a d^2&=6336 \\ \Rightarrow 3(12)^3+6(12) d^2&=6336 \text { as } a=12 \\ \Rightarrow 3(1728)+72 d^2&=6336 \\ \Rightarrow 72 d^2&=6336-5184=1152 \\ \Rightarrow d^2&=16= \pm 4\end{align} When $a=12$ and $d=4$ then the numbers are
\begin{align} a-d&=12-4=8, a=12 \text { and } \\ a+d&=12+4=16 .\end{align} When $a=12$ and $d=-4$ then the numbers are
\begin{align}a-d&=12-(-4)=16, a=12, \text { and } \\ a+d&=12+(-4)=8\\ & 8,12,16; 16,12,8\end{align}