Question 2 Exercise 4.3

Solutions of Question 2 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.

Some of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that is missing: $a_1=2, n=17, d=3$.

Given: $a_1=2, n=17, d=3$
We need to find $a_{17}$ and $S_{17}$. As we know $$a_{n}=a_1+(n-1)d.$$ Thus $$a_{17}=2+(17-1)(3)=50.$$ Also $$S_n=\dfrac{n}{2}[a_1+a_n]$$ Thus $$\begin{align}S_{17}&=\dfrac{17}{2}(a_1+a_17) \\ &=\dfrac{17}{2}(2+50)=442.\end{align}$$ Hence $a_{17}=50$ and $S_{17}=442$.

Some of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that are missing $a_1=-40, S_{21}=210$.

Given: $a_1=-40$ and $S_{21}=210$.
So we have $n=21$ and we have to find $a_{21}$ and $d$. As $$\begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\ \implies &210=\dfrac{21}{2}(-40+a_{21}) \\ \implies &-40+a_{21}=\dfrac{2 \times 210}{21}=20 \\ \implies &a_{21}=20+40=60.\end{align}$$ Also $a_{21}=a_1+20 d$, then
$$\begin{align}&20d=60-(-40)=100 \\ \implies &d=\dfrac{100}{20}=5 \end{align}$$ Hence $a_{21}=60$, $d=5$ and $n=21$.

Some of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that are missing $a_1=-7, d=8, S_n=225$.

Given: $a_1=-7, d=8, S_n=225$, we have to find $n$ and $a_n$. We know that $$S_n=\dfrac{n}{2}[2 a_1+(n-1) d].$$ Thus, we have $$\begin{align} & 225=\dfrac{n}{2}[2 \cdot(-7)+(n-1) \cdot 8], \\ \implies & n[-14+8(n-1)]=2 \times 225 \\ \implies & -14 n+8 n(n-1)=450 \\ \implies & 8 n^2-8 n-14 n=450 \\ \implies & 8 n^2-22 n-450=0 \\ \implies & 4 n^2-11 n-225=0\end{align}$$ This is quadratic equation with $a=4, b=-11$ and $c=-225$, then $$\begin{align}n&=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ n&=\dfrac{11 \pm \sqrt{(-11)^2-4(4)(-225)}}{2.4} \\ \implies n&=\dfrac{11 \pm \sqrt{121+3600}}{8} \\ \implies n&=\dfrac{11 \pm \sqrt{3721}}{8}\\ &=\dfrac{11 \pm 61}{8} \\ \implies n&=\dfrac{11+61}{8} \text { or } n=\dfrac{11-61}{8} \\ \implies n&=9 \text { or } n=-\dfrac{50}{8}\end{align}$$ Since $n$ cannot be negative or in fraction, thus $n=9$. Now $$a_9=a_1+8 d=-7+8(8)=57.$$ Hence $n=9$ and $a_{9}=57$.

Some of the components $a_1, a_n, n, d$ and $S_n$ are given. Find the one that are missing: $a_n=4, S_{15}=30$.

Given: $a_n=4, S_{15}=30$.
Thus we have $n=15$ and we have to find $a_1$ and $d$. We know that $$S_n=\dfrac{n}{2}[a_1+a_n],$$ then we have $$\begin{align}&S_{15}=\dfrac{15}{2}[a_1+a_{15}]\\ \implies &\dfrac{15}{2}[a_1+4]=30 \\ \implies &a_1+4=\dfrac{30 \times 2}{15}=4 \\ \implies &a_1=4-4=0. \end{align}$$ Also $$\begin{align}&a_{15}=a_1+14d \\ \implies &4=0+14d \\ \implies &d=\dfrac{4}{14}=\dfrac{2}{7}.\end{align}$$ Hence $a_1=0$, $n=15$ and $d=\dfrac{2}{7}$.

< Question 1 Question 3 & 4 >