Sum of the first n terms of an arithmetic progression : Problems and Solutions (Part 2)

Sum of the first $n$ terms of an arithmetic progression $\left(S_{n}\right)$

$S_{n}=u_{1}+u_{22}+u_{3}+\ldots u_{n-1}+ u_{n}$

$S_{n}=$ sum of the first $n$ terms

If $u_{1}=a$ and $u_{n}=l$ then

$\begin{array}{|l|} \hline S_{n}=\dfrac{n}{2}(a+l)\\ \quad \text{(or) }\\ \quad S_{n}=\dfrac{n}{2}\{2 a+(n-1) d\}\\ \hline \end{array}$

$a=$ first term

$l=$ last term or $n^{\text {th }}$ term

$d=$ common difference

Arithmetic Series နှင့်ဆိုင်သော သင်ရိုးပြင်ပ ဉာဏ်စမ်းမေးခွန်းများ ကို စုစည်းပေးထားပြီး အဖြေနှင့်တကွ လေ့လာနိုင်ရန် ဖြစ်ပါသည်။ Part (1) ကို ဒီနေရာမှာ (Click Here) ရေးခဲ့ပြီး ဖြစ်ပါသည်။

Exercises

1. If the $n^{\text {th }}$ term of an A.P. is $p$, show that the sum of first $(2 n-1)$ terms of the A.P. is $(2 n-1) p$.





Let the first term and the conmon difference of given A.P.be $a$ and $d$ respectively. By the problem, $\begin{aligned} u_{n}&=p \\\\ a+(n-1) d&=p\\\\ S_{2 n-1}&=\dfrac{2 n-1}{2}\left\{2 a+((2 n-1)-1) d\right\} \\\\ S_{2 n-1}&=\dfrac{2 n-1}{2}\left\{2 a+(2 n-2) d\right\} \\\\ S_{2 n-1}&=\dfrac{2 n-1}{2}\left\{2(a)+2(n-1) d\right\} \\\\ S_{2 n-1}&=\dfrac{2 n-1}{2}\left\{2(a+(n-1) d\right\} \\\\ S_{2 n-1}&=\dfrac{2 n-1}{2}\left\{2 p\right\} \\\\ S_{2 n-1}&=(2 n-1) p \end{aligned}$



2. The digits of a positive integer having three digits are in A.P. The sum of the digits is 15 and the number obtained by reversing the digits is 594 less than the original number. Find the number.





Let the hundred's digit, ten's digit and one's digit of given number be $x, y$ and $z$ respectively. $\therefore$ Given number $=100 x+10 y+z$ $\therefore x, y, z$ are in A.P. Let tho common differenee be $d$. $\therefore y=x+d, z=x+2 d$ By the problen, $\begin{aligned} x+y+z &=15 \\\\ 3 x+3 d &=15 \\\\ x+d &=5 \\\\ x &=5-d \\\\ \text { reversed-digit number } &=100 z+10 y+x \\\\ \text { By the problem, }\hspace{4cm} & \\\\ 100 x+10 y+z-100 z-10 y-x &=594 \\\\ 99 x-99 z &=594 \\\\ x-z &=6 \\\\ 5-d-(5-d+2 d) &=6 \\\\ 5-d-5+d-2 d &=6 \\\\ \therefore d &=-3 \\\\ \therefore x &=5-(-3) \\\\ &=8\\\\ \therefore y=5 \text { and } z &=2 \\\\ \therefore \text { Given number } &=100 x+10 y+z \\\\ &=100(8)+10(5)+2 \\\\ &=852 \end{aligned}$



3. In an A.P., the last three terms are $45,51$ and $57$. Which of the numbers $-180$, $-210$ and $-288$ can be the sum of that A.P. and what is the first term of that A.P.





Let the first term and the conmon difference of given A.P.be $a$ and $d$ respectively. $\begin{aligned} \therefore\ d &=57-51=6 \\\\ u_{n} &=a+(n-1) d \\\\ 57 &=a+(n-1) 6 \\\\ a &=57-6(n-1) \\\\ &=63-6 n\\\\ S_{n} &=\dfrac{n}{2}(a+57) \\\\ &=\dfrac{n}{2}(63-6 n+57) \\\\ &=\dfrac{n}{2}(120-6 n) \\\\ &=60 n-3 n^{2} \\\\ \text { When } 60 n-3 n^{2}&=-180 \\\\ & n^{2}-20 n=60 \\\\ & n^{2}-20 n+100=160 \\\\ &(n-10)^{2}=160 \\\\ & n=10+4 \sqrt{10} \notin \mathbb{N}.\\\\ \therefore S_{n} &=-180 \text { is impossible. } \\\\ \text { When } 60 n-3 n^{2} &=-210 \\\\ n^{2}-20 n &=70 \\\\ n^{2}-20 n+100 &=170 \\\\ (n-10)^{2} &=170 \\\\ n &=10+\sqrt{170} \notin \mathbb{N} \\\\ \therefore S_{n}&=-210 \text { is impossible. } \\\\ \text { When } 60 n-3 n^{2} & =-288 \\\\ n^{2}-20 n & =96 \\\\ n^{2}-20 n+100 & =196 \\\\ (n-10)^{2} & =196 \\\\ n & =10+14 \\\\ & =24 \in\mathbb{N} \\\\ \therefore S_{n}&=-288 \text { is possible. } \\\\ \therefore a & =63-6(24) \\\\ & =-81 \end{aligned}$



4. If $S_{n}$ denotes the sum of $n$-terms of an A.P. whose common difference is $d$. Show that $d=S_{n}-2 S_{n-1}+S_{n-2}$.





$\begin{aligned} \text{In an A.P.,}&\\\\ &S_{n}-2 S_{n-1}+S_{n-2} \\\\ =\ & S_{n}-S_{n-1}-S_{n-1}+S_{n-2} \\\\ =\ &\left(S_{n}-S_{n-1}\right)-\left(S_{n-1}-S_{n-2}\right) \\\\ =\ & u_{n}-u_{n-1}\quad \color{red}{\left(\because u_{n}=S_{n}-S_{n-1}\right)} \\\\ =\ & d \end{aligned}$



5. Show that the sum of first $n$ even natural numbers is equal to $\left(1+\dfrac{1}{n}\right)$ times the sum of the first $n$ odd natural numbers.





The sequence of even natural number are $2,4,6,8, \ldots$. It is an A.P., with $a=2$ and $d=2$. Let the sum of $n$ even natural numbers be $S_{\text{even}}$. $\begin{aligned} S_{\text{even}} &=\dfrac{n}{2}\{2 a+(n-1) d\} \\\\ &=n(2+n-1) \\\\ &=n(n+1)\\\\ \end{aligned}$ The sequence of even natural number are $1,3,5,7, \ldots$. It is an A.P., with $a=1$ and $d=2$. Let the sum of $n$ odd natural numbers be $S_{\text{odd}}$. $\begin{aligned} S_{\text{odd}} &=\dfrac{n}{2}\{2 a+(n-1) d\} \\\\ &=n(1+n-1) \\\\ &=n^2\\\\ \dfrac{S_{\text{even}}}{S_{\text{odd}}} &=\dfrac{n(n+1)}{n^{2}} \\\\ &=\dfrac{n+1}{n} \\\\ &=1+\dfrac{1}{n} \\\\ \therefore \ S_{\text{even}} &=\left(1+\frac{1}{n}\right) S_{\text{odd}} \end{aligned}$



6. Find the sum of the A.P., whose terms are $1,5,9,13, \ldots, 605$. If every fourth term of the A.P. (i.e. 13,29 , etc.) is taken out, find the sum of the remaining terms.





$\begin{aligned} 1,5,9,13, & \ldots .605 \text { is an } A P . \\\\ a=1, d &=5-1=4 \\\\ u_{n} &=605 \\\\ 1+(n-1) 4 &=605 \\\\ 4(n-1) &=604 \\\\ n-1 &=151 \\\\ n &=152 \\\\ \therefore\ S_{n} &=\dfrac{n}{2}(a+1) \\\\ S_{152} &=\dfrac{152}{2}(1+605) \\\\ &=46056\\\\ \end{aligned}$ Every fourth terms of given A.P. are $13,29,45, \ldots$. It is also an A.P with $a=13, d=16$. Let the last term be $u_n$. $\begin{aligned} u_{n} & \leq 605 \\\\ 13+(n-1) 16 & \leq 605 \\\\ (n-1) 16 & \leq 592 \\\\ n-1 & \leq 37 \\\\ n & \leq 38 \end{aligned}$ $\therefore$ The last term of new A.P. is 605 and there are 38 terms in the new A.P. Let the sum of all terms in the new A.P. be $S_{\text {new }}$. $\begin{aligned} \therefore\ S_{\text {new }} &=\dfrac{38}{2}(13+605) \\\\ &=11742 \\\\ \therefore\ \text { remaining sum } &=46056-11742 \\\\ &=34314 \end{aligned}$



7. Find the sum of all two-digit numbers which when divided by $4$ , yields $1$ as remainder.





All two-digit numbers which when divided by 4 yields 1 as remainder are $13,17,21, \ldots, 97 .$ It is an A.P. with $a=13$ and $d=4$. Let $u_{n}=97$, then $\begin{aligned} 13+(n-1) 4 &=97 \\\\ (n-1) 4 &=84 \\\\ n-1 &=21 \\\\ n &=22\\\\ S_{n} &=\dfrac{n}{2}(a+l) \\\\ S_{22} &=\dfrac{22}{2}(13+97) \\\\ &=11(110) \\\\ &=1210 \end{aligned}$



8. Find the sum of first $24$ terms of the A.P. $u_{1}, u_{2}, u_{3}, \ldots, u_{n}$ if it is known that $u_{1}+u_{5}+u_{10}+u_{15}+u_{20}+u_{24}=225 .$





$\begin{aligned} u_{1},\ u_{2},\ u_{3}, \ \ldots, u_{n} \text { is an A.P. }& \\\\ u_{1}+u_{5}+u_{10}+u_{15}+u_{20}+u_{24} &=225 \\\\ \dfrac{6}{2}\left(u_{1}+u_{24}\right) &=225 \\\\ a+a+23 d &=75 \\\\ 2 a+23 d &=75 \\\\ S_{n} &=\dfrac{n}{2}\{2 a+(n-1) d\} \\\\ S_{2 y} &=\dfrac{24}{2}\{2 a+23 d\} \\\\ &=12\{75\} \\\\ &=900 \end{aligned}$



9. If in an A.P., the sum of $m$ terms is equal to $n$ and the sum of $n$ terms is equal to $m$, prove that the sum of $(m+n)$ terms is $-(m+n)$.





$\begin{aligned} \text{In an A.P.,}\\\\ S_{m} &=n \\\\ \dfrac{m}{2}\{2 a+(m-1) d\} &=n \\\\ 2 a m+\left(m^{2}-m\right) d &=2 n \ldots(1) \\\\ S_{n} &=m \\\\ \dfrac{n}{2}\{2 a+(n-1) d\} &=m \\\\ 2 a n+\left(n^{2}-n\right) d &=2 m \ldots(2)\\\\ \end{aligned}$ $\begin{aligned} \text { By (1) - (2), }\hspace{6cm}& \\\\ 2 a(m-n) +\left(m^{2}-m-n^{2}+n\right) d&=2(n-m) \\\\ 2 a(m-n) +\left[m^{2}-n^{2}-(m-n)\right] d&=-2(m-n) \\\\ 2 a(m-n) +[(m-n)(m+n)-(m-n)] d&=-2(m-n) \\\\ \end{aligned}$ $\begin{aligned} \text { Dividing both sides with } & m-n , \\\\ 2 a+(m+n-1) d&=-2 \\\\ S_{m+n}&=\dfrac{m+n}{2}\{2 a+(m+n-1) d\} \\\\ &=\dfrac{m+n}{2}(-2) \\\\ &=-(m+n) \end{aligned}$



10. In an A.P., if $S_{n}=n^{2} p$ and $S_{m}=m^{2} p, m \neq n$, prove that $S_{p}=p^{3}$.





$\begin{aligned} \text { In an A.P., }\quad\quad& \\\\ S_{n} &=n^{2} p \\\\ \dfrac{n}{2} \left\{ 2 a+(n-1) d \right\} &=n^{2} p \\\\ 2 a+(n-1) d &=2 n p\ldots(1) \\\\ S_{m} &=m^{2} p \\\\ \dfrac{m}{2}\left\{ 2 a+(m-1) d \right\} &=m^{2} p \\\\ 2 a+(m-1) d &=2 m p \ldots(2)\\\\ \text{By } (1)-(2),\quad\quad & \\\\ (n-1-m+1) d &=2 p(n-m) \\\\ (n-m) d &=2 p(n-m) \\\\ d &=2 p\\\\ \text {Substituting } d&=2p \text { in } (1),\\\\ 2 a+(n-1) 2 p &=2 n p \\\\ 2 a+2 n p-2 p &=2 n p \\\\ 2 a &=2 p \\\\ a &=p \\\\ \end{aligned}$ $\begin{aligned} \therefore\ S_{p} &=\dfrac{p}{2}\left\{ 2 a+(p-1) d \right\} \\\\ &=\dfrac{p}{2}\left\{ 2 p+(p-1) 2 p \right\} \\\\ &=p \left\{ p+p^{2}-p \right\} \\\\ &=p^{3} \end{aligned}$



11. If $S_{1}, S_{2}, S_{3}, \ldots, S_{m}$ are the sum of $n$ terms of $m$ arithmetic progressions whose first terms are $1,2,3, \ldots, m$ and the common differences are $1,3,5, \ldots,(2 m-$1), respectively. Show that $S_{1}+S_{2}+S_{3}+\ldots+S_{m}=\dfrac{m n}{2}(m n+1)$.





$\begin{aligned} &\text{ By the problem, }\\\\ &S_{1}=\dfrac{n}{2}\left\{2+(n-1)\right\} \\\\ &S_{2}=\dfrac{n}{2}\left\{4+(n-1) 3\right\} \\\\ &S_{3}=\dfrac{n}{2}\left\{6+(n-1) 5\right\} \\\\ &\vdots \\\\ &S_{m}=\dfrac{n}{2}\left\{2 m+(n-1)(2 m-1)\right\}\\\\ \end{aligned}$ $\quad\ S_{1}+S_{2}+S_{3}+\ldots+S_{m}\\\\ $ $=\dfrac{n}{2}\{2+4+6+\cdots+2 m+(n-1)(1+3+5+\cdots(2 m-1)\}\\\\ $ Let $S_{a}=2+4+6+\cdots+2 m\\\\ $ Since $4-2=2$ and $6-4=2,\\\\\ $ $\text{the terms are in A.P.}\\\\ $ $\therefore\ S_{a}=\dfrac{m}{2}\{2(2)+(m-1) 2\}\\\\ $ $\hspace{2cm}=m(m+1)\\\\ $ $\text { Let } S_{b}=1+3+5+\ldots+(2 m-1)\\\\ $ $\text { Since } 3-1=2 \text { and } 5-3=2,\\\\ $ $\text{the terms are in A.P.}\\\\ $ $\begin{aligned} S_{b} &=\dfrac{m}{2}\{2+(m-1) 2\} \\\\ &=n^{2} \\\\ S_{1} & +S_{2}+S_{3}+\cdots+S_{m} \\\\ &=\dfrac{n}{2}\left\{S_{a}+(n-1) S_{b}\right\} \\\\ &=\dfrac{n}{2}\left\{m(m+1)+(n-1) m^{2}\right\} \\\\ &=\dfrac{m n}{2}(m+1+m n-m) \\\\ &=\dfrac{m n}{2}(m n+1) \end{aligned}$



12. If $S_{n}$ denotes the sum of the first $n$ terms of an A.P., prove that $\dfrac{S_{3 n}-S_{n-1}}{S_{2 n}-S_{2 n-1}}=2 n+1$.





Let the $a$ denotes the first term and $d$ denotes the common difference. $\begin{aligned} &\\\\ \therefore S_{n} &=\dfrac{n}{2}\{2 a+(n-1) d\} \\\\ S_{3 n-S_{n-1}} &=\dfrac{3 n}{2}\{2 a+(3 n-1) d\}-\dfrac{n-1}{2}\{2 a+(n-2) d\} \\\\ &=\dfrac{3 n}{2}(2 a)+\dfrac{3 n}{2}(3 n-1) d-\left(\dfrac{n-1}{2}(2 a)-\dfrac{(n-1)(n-2)}{2}\right) d \\\\ &=(3 n-n+1) a+\left(\dfrac{9 n^{2}-3 n-n^{2}+3 n-2}{2}\right) d \\\\ &=(2 n+1) a+\left(4 n^{2}-1\right) d \\\\ &=(2 n+1) a+(2 n+1)(2 n-1) d \\\\ &=(2 n+1)\{a+(2 n-1) d\}\\\\ S_{2 n}-S_{2 n-1} &=\dfrac{2 n}{2}\{2 a+(2 n-1) d\}-\dfrac{2 n-1}{2}\{2 a+(2 n-2) d\} \\\\ &=2 n a+\left(2 n^{2}-n\right) d-(2 n-1) a-(2 n-1)(n-1) d \\\\ &=(2 n-2 n+1) a+\left(2 n^{2}-n-2 n^{2}+3 n-1\right) d \\\\ &=a+(2 n+1) d \\\\ \dfrac{S_{3 n}-S_{n-1}}{S_{2 n}-S_{2 n-1}} &=\dfrac{(2 n+1)\{a+(2 n-1) d\}}{a+(2 n-1) d}\\\\ &=2n+1 \end{aligned}$



13. If the sum of the first $2 n$ terms of the A.P. $2,5,8, \ldots$ is equal to the sum of the first $n$ terms of the A.P. $57,59,61, \ldots$, find $n$.





$\begin{aligned} 2,\ 5,\ 8,\ \ldots & \text { is an A.P. } & \\\\ a=2,\ d &=3 \\\\ S_{2 n} &=\dfrac{2 n}{2}\{2 a+(2 n-1) d\} \\\\ &=n\{2(2)+(2 n-1) 3\} \\\\ &=n(4+6 n-3) \\\\ &=n(6 n+1)\\\\ 57,\ 59,\ 61,\ \ldots, & \text { is an A.P. } \\\\ a= 57,\ d &= 2 \\\\ S_{n} &=\dfrac{n}{2}\{2 a+(n-1) d\} \\\\ &=\dfrac{n}{2}\{2(57)+(n-1) 2\} \\\\ &=n(57+n-1) \\\\ &=n(n+56)\\\\ \text{By the problem,}&\\\\ n(6 n+1) &=n(n+56) \\\\ 6 n+1 &=n+56 \\\\ 5 n &=55 \\\\ n &=11 \end{aligned}$



14. An arithmetic series has $20$ terms. The $n^{\text{th}}$ term is $u_{n}$ and $S_{n}$ is the sum of the first $n$ terms of this series. If $S_{5}=85$, find $u_{3}$. Given further that $S_{17}=35 u_{3}$, evaluate the sum of all the terms of this series.





Let $a$ denotes the finst term and $d$ denotes the common difference. $\begin{aligned} &\\ S_{n}&=\dfrac{n}{2}\{2 a+(n-1) d\} \\\\ S_{5}&=85 \quad \text { (given) } \\\\ \dfrac{5}{2}\{2 a+4 d\}&=85 \\\\ a+2 d&=17 \ldots(1)\\\\ \end{aligned}$ Since $u_{3}=a+2 d, u_{3}=17.\\ $ $\begin{aligned} S_{17}&=35 u_{3} \\\\ \dfrac{17}{2}\{2 a+16 d\}&=35(17) \\\\ a+8 d&=35\ldots(2) \\\\ \text{By } (2)-(1),& \\\\ 6d&=18 \\\\ d&=3 \\\\ S_{20}&=\dfrac{20}{2}\{2 a+19 d\} \\\\ &=10\{2(a+8 d)+3 d\} \\\\ &=10\{2(35)+9\} \\\\ &=790 \end{aligned}$



15. In an arithmetic series, $u_{8}=26$ and $S_{5}=205$. Calculate the smallest positive term of the series.





In an arithmetic series, $\begin{aligned} &\\ u_{8}&=26\ldots(1) \\\\ a+7 d&=26 \\\\ S_{5}&=205 \\\\ \dfrac{5}{2}\{2 a+4 d\}&=205 \\\\ a+2 d=41\ldots(1)\\\\ \text{By } (1)-(2),& \\\\ 5d &=-15 \\\\ d &=-3\\\\ \end{aligned}$ Substituting $d=-3$ in equation (2), $\begin{aligned} &\\ &a-6=41 \\\\ &a=47 \\\\ \end{aligned}$ Let the smallest positive term be $u_{n}\\\\ $. $\begin{aligned} u_{n} &>0 \\\\ a+(n-1) d &>0 \\\\ 47+(n-1)(-3) &>0 \\\\ 47-3 n+3 &>0 \\\\ 3 n & < 50 \\\\ n & <16.67 \\\\ \therefore\ n&=16\\\\ \end{aligned}$ The smallest positive term is $u_{16}$.



16. The first term of an arithmetic series is $5$ and the common difference is $4$ The $n^{\text {th }}$ term is $u_{n}$ and $S_{n}$ is the sum of the first $n$ terms of this series. Show that $S_{n}=n(2 n+3)$. Find the value of $u_{n}$, given that $S_{n}=779$.





In an arithmetic series, $\begin{aligned} &\\ a &=5, d=4 \\\\ S_{n} &=\dfrac{n}{2}\{2 a+(n-1) d\} \\\\ &=\dfrac{n}{2}\{2(5)+(n-1) 4\} \\\\ &=n(5+2 n-2) \\\\ &=n(2 n+3)\\\\ S_{n} &=779 \\\\ n(2 n+3) &=779 \\\\ 2 n^{2}+3 n-779 &=0 \\\\ (2 n+41)(n-19) &=0 \\\\ n=-\dfrac{41}{2} \notin \mathbb{N} \text { or } n &=19 \\\\ \therefore\ u_{n} &=u_{19} \\\\ &=a+18 d \\\\ &=5+18(4) \\\\ &=77 \end{aligned}$



17. The first and last terms of an A.P. are $a$ and $l$ respectively. If $d$ is the common difference and $S$ is the sum of all the terms of the A.P., then show that $d=\dfrac{l^{2}-a^{2}}{2 S-l-a}$.





$\text{In an A.P.}\\\\ $ $\text{first term }=a\\\\ $ $\text{Last term }=l\\\\ $ $\text{common difference }=d\\\\ $ $\text{sum of all terms }=S\\\\ $ Let the number of terms in the given A.P. be $n\\\\ $. $\begin{aligned} \therefore\ l&=u_{n}=a+(n-1) d \\\\ \therefore\ (n-1) d&=l-a\ldots(1) \\\\ S=S_{n}&=\dfrac{n}{2}(a+l) \\\\ \therefore\ n&=\dfrac{2S}{a+1}\ldots(2)\\\\ \text { Substituting } n &=\dfrac{2 S}{a+l} \text { in equation }(1), \\\\ \left(\dfrac{2S-l-a}{l+a}\right) d &=l-a \\\\ \therefore\ d &=\dfrac{l^{2}-a^{2}}{2S-l-a} \end{aligned}$

စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!