Sum of the first n terms of an arithmetic progression : Problems and Solutions (Part 1)

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

Exercises

1. The sum to $n$ terms of an A.P. is $35$. The common difference is $2$ and the sum to $2 n$ terms is 120 . Find the first term.





$\begin{aligned} \text{In an A.P.,} &\\\\ d &=2, \\\\ S_{n} &=35 \\\\ \dfrac{n}{2}\{2 a+(n-1) 2\} &=35 \\\\ \dfrac{n}{2} \times 2\{a+n-1\} &=35 \\\\ a n+n^{2}-n &=35\ldots(1) \\\\ S_{2 n} &=120 \\\\ \dfrac{2 n}{2}\{2 a+(2 n-1) 2\} &=120 \\\\ n \times 2\{a+2 n-1\} &=120 \\\\ a n+2 n^{2}-n &=60\ldots(2) \\\\ (2)-(1), & \\\\ n^{2} &=25 \\\\ n &=5 \\\\ \text { Substituting } n &=5 \text { in equation (1), } \\\\ 5 a+25-5 &=35 \\\\ 5 a &=15 \\\\ a &=3 \end{aligned}$



2. An A.P., with first term 8 and common difference $d$, consists of 101 terms. Given that the sum of the last three terms is 3 times the sum of the first three terms, find the value of $d$.





In an AP, first term $=a=8$ common difference $=d$ By the problem, $\begin{aligned} u_{99}+u_{100}+u_{101} &=3\left(u_{1}+u_{2}+u_{3}\right) \\\\ a+98 d+a+99 d+a+100 d &=3(a+a+d+a+2 d) \\\\ 3 a+297 d &=9 a+9 d \\\\ 288 d &=6 a \\\\ d &=\dfrac{6}{288}(8) \\\\ &=\dfrac{1}{6} \end{aligned}$



3. The sum of the first $n$ terms of an A.P. $3,5 \dfrac{1}{2}, 8, \ldots$ is equal to the $2 n^{\text {th }}$ term of the A.P $16 \dfrac{1}{2}, 28 \dfrac{1}{2}, 40 \dfrac{1}{2}, \ldots .$ Calculate the value of $n$.





$\begin{aligned} 3,\ 5\dfrac{1}{2},\ & 8,\ \ldots \text { is an A.P. } \\\\ a &=3 \\\\ d &=5 \dfrac{1}{2}-3 \\\\ &=2 \dfrac{1}{2} \\\\ &=\dfrac{5}{2} \\\\ S_{n} &=\dfrac{n}{2}\{2 a+(n-1) d\} \\\\ &=\dfrac{n}{2}\left\{6+(n-1) \dfrac{5}{2}\right\} \\\\ &=\dfrac{n}{4}\{12+5 n-5\} \\\\ &=\dfrac{5 n^{2}+7 n}{4}\\\\ 16 \dfrac{1}{2},\ & 28 \dfrac{1}{2},\ 40 \dfrac{1}{2},\ \ldots \text { is an A.P. } \\\\ a&=16 \dfrac{1}{2}\\\\ &=\dfrac{33}{2} \\\\ d &=28 \dfrac{1}{2}-16 \dfrac{1}{2}\\\\ &=12 \\\\ u_{2 n} &=a+(2 n-1) d \\\\ &=\dfrac{33}{2}+(2 n-1) 12 \\\\ &=24 n+ \dfrac{9}{2}\\\\ \text{By the } & \text{problem},\\\\ \dfrac{5 n^{2}+7 n}{4}&=24 n+\dfrac{9}{2}\\\\ 5 n^{2}+7 n &=96 n+18 \\\\ 5 n^{2}-89 n-18 &=0 \\\\ (5 n+1)(n-18) &=0 \\\\ n=-\dfrac{1}{5} \text { or } n &=18 \\\\ \text { Since }-\dfrac{1}{5} & \notin N, \\\\ n &=18 \end{aligned}$



4. The sum of the first $n$ terms of a certain sequence is given by $S_{n}=n^{2}+2 n$. Find the first 3 terms of the sequence and express the $n^{\text {th }}$ term in terms of $n$.





$\begin{aligned} S_{n} &=n^{2}+2 n \\\\ S_{1} &=1^{2}+2(1)=3 \\\\ S_{2} &=2^{2}+2(2)=8 \\\\ S_{3} &=3^{2}+2(3)=15 \\\\ \text {Since } u_{n} &=S_{n}-S_{n-1}, \\\\ u_{1} &=S_{1}=3 \\\\ u_{2} &=S_{2}-S_{1}=5\\\\ u_{3} &=S_{3}-S_{2}=15-8=7\\\\ \text { Since }\\\\ u_{2}-u_{1} &=5-3=2 \text { and } \\\\ u_{3}-u_{2} &=7-5=2,\\\\ \text{the terms } & \text{are in A.P., with}\\\\ a=3 \text { and } & d =2 \\\\ u_{n} &=a+(n-1) d \\\\ &=3+(n-1) 2 \\\\ &=3+ 2n-2 \\\\ &=2 n+1 \end{aligned}$



5. If the sum of $n$ terms of a certain sequence is $2 n+3 n^{2}$, find the $n^{\text {th }}$ term. Hence show that this sequence is an arithmetic progression.





$\begin{aligned} S_{n} &=2 n+3 n^{2} \\\\ u_{n} &=S_{n}-S_{n-1} \\\\ &=2 n+3 n^{2}-2(n-1)-3(n-1)^{2} \\\\ &=2 n+3 n^{2}-(2 n-2)-3\left(n^{2}-2 n+1\right) \\\\ &=2 n+3 n^{2}-2 n+2-3 n^{2}+6 n-3 \\\\ &=4 n-1 \\\\ u_{n-1} &=S_{n-1}-S_{n-2} \\\\ &=2(n-1)+3(n-1)^{2}-2(n-2)-3(n-2)^{2} \\\\ &=2 n-2+3\left(n^{2}-2 n+1\right)-2 n+4-3\left(n^{2}-4 n+4\right) \\\\ &=2 n-2+3 n^{2}-6 n+3-2 n+4-3 n^{2}+12 n-12 \\\\ &=4 n-7\\\\ u_{n}-u_{n-1} &=4 n-1-(4 n-7) \\\\ &=6 \end{aligned}$ Since the difference of two consicutive terms is constant, the sequence is an A.P.



6. The sum of $n$ terms of two arithmetic progressions are in the ratio $(3 n+8) \vdots(7 n+$ 15). Find the ratio of their $12^{\text {th }}$ terms.





Let the st term and the common difference of $1^{\text{st}}$ A.P be $a$ and $d$ and those of $2^{\text{nd}}$ A.P be $A$ and $D$, For $1^{\text{st}}$ A.P., $S_{n}=\dfrac{n}{2}\{2 a+(n-1) d\}$ For $2^{\text{nd}}$ A.P., $S_{n}=\dfrac{n}{2}\{2 A+(n-1) D\}$ By the problem, $\begin{aligned} \dfrac{\dfrac{n}{2}\{2 a+(n-1) d\}}{\dfrac{n}{2}\{2 A+(n-1) D\}}&=\dfrac{3 n+8}{7 n+15}\\\\ \text { When } n=23,\hspace{2 cm} & \\\\ \dfrac{2 a+22 d}{2 A+22 D}&=\dfrac{77}{176}\\\\ \dfrac{a+11d}{A+11 D}&=\dfrac{7}{16}\\\\ \therefore\ \dfrac{12^{\text {th }} \text { term of } 1^\text {st} \text {A.P}}{12^{\text {th }} \text { term of } 2^\text{nd} \text { A.P}}&=\dfrac{7}{16} \end{aligned}$



7. An A.P. contains $30$ terms. Given that the $10^{\text {th }}$ term is $21$ and that the sum of the last $10$ terms is $675$ , find the sum of the first $10$ terms.





$\begin{aligned} \text{In an A.P.,}\hspace{3 cm}&\\\\ u_{10} &=21 \\\\ a+9 d &=21\ldots(1) \\\\ u_{21}+u_{22}+u_{23}+\ldots+u_{30} &=675 \\\\ \dfrac{10}{2}\left\{u_{21}+u_{30}\right\} &=675 \\\\ 5\{a+20 d+a+29 d\} &=675 \\\\ 2 a+49 d &=135\ldots(2)\\\\ (2) \times 1 \Rightarrow 2 a+49 d&=135 \\\\ (1) \times 2 \Rightarrow 2 a+18 d&=42\\\\ \text{By subtraction,}\hspace{2 cm}&\\\\ 31 d &=93 \\\\ d &=3\\\\ \end{aligned}$ $\begin{aligned} \text{Substituting } d&=3 \text{ in equation } (1),\\\\ a+27 &=21 \\\\ a &=-6\\\\ S_{10}&=\dfrac{10}{2}\{2 a+9 d\} \\\\ &=5\{-12+27\} \\\\ &=75 \end{aligned}$



8. If $S_{1}, S_{2}, S_{3}$ be sums to $n, 2 n, 3 n$ terms of an arithmetic progression, Show that $S_{3}=3\left(S_{2}-S_{1}\right)$.





Let the first term and the common difference of given A.P. be $a$ and $d$ respectively. $\begin{aligned} \therefore\ S_{1} &=\dfrac{n}{2}\{2 a+(n-1) d\} \\\\ S_{2} &=\dfrac{2 n}{2}\{2 a+(2 n-1) d\} \\\\ S_{3} &=\dfrac{3 n}{2}\{2 a+(3 n-1) d\} \\\\ S_{2}-S_{1} &=\dfrac{2 n}{2}\{2 a+(2 n-1) d\}-\dfrac{n}{2}\{2 a+(n-1) d\} \\\\ &=\dfrac{n}{2}\{4 a+4 n d-2 d-2 a-n d+d\} \\\\ &=\dfrac{n}{2}\{2 a+3 n d-d\} \\\\ &=\dfrac{n}{2}\{2 a+(3 n-1) d\} \\\\ 3\left(S_{2}-S_{1}\right) &=\dfrac{3 n}{2}\{2 a+(3 n-1) d\} \\\\ \therefore\ S_{3} &=3\left(S_{2}-S_{1}\right) \end{aligned}$



9. In an A.P., if $p^{\text {th }}$ term is $\dfrac{1}{q}$ and $q^{\text {th }}$ term is $\dfrac{1}{p}$, prove that the sum of first $p q$ terms is $\dfrac{1}{2}(p q+1)$, where $p \neq q$.





$\begin{aligned} \text{In an A.P,}&\\\\ u_{p} &=\dfrac{1}{q} \\\\ a+(p-1) d &=\dfrac{1}{q}-(1) \\\\ u_{q} &=\dfrac{1}{p} \\\\ a+(q-1) d &=\dfrac{1}{p}\\\\ (1)-(2), & \\\\ (p-q) d &=\dfrac{1}{q}-\dfrac{1}{p} \\\\ (p-q) d &=\dfrac{p-q}{p q} \\\\ d &=\dfrac{1}{p q}\\\\ \end{aligned}$ $\begin{aligned} \text{Substituting } d&=\dfrac{1}{q-p} \text{ in equation} (1),\\\\ a+\dfrac{p-1}{p q} &=\dfrac{1}{q} \\\\ a &=\dfrac{1}{q}-\dfrac{p-1}{p q} \\\\ &=\dfrac{p-p+1}{p q} \\\\ &=\dfrac{1}{p q}\\\\\ \therefore\ S_{p q} &=\dfrac{p q}{2}\{2 a+(p q-1) d\} \\\\ &=\dfrac{p q}{2} \left\{\dfrac{2}{p q}+(p q-1) \dfrac{1}{p q}\right\} \\\\ &=1+\dfrac{1}{2}(p q-1) \\\\ &=\dfrac{1}{2} p q+1-\dfrac{1}{2}=\dfrac{1}{2} p q+\dfrac{1}{2} \\\\ &=\dfrac{1}{2}(p q+1) \end{aligned}$ Thus, the sum of first $p q$ terms of the A.P. is $\dfrac{1}{2}(p q+1)$



10. An arithmetic progression has third term $90$ and fifth term $80$.
    (a) Find the first term and the common difference.
    (b) Find the value of $m$ given that the sum of the first $m$ terms is equal to the sum of the first $(m+1)$ terms.
    (c) Find the value of $n$ given that the sum of the first $n$ terms is zero.





$\begin{aligned} \text{ (a) In an A.P.,} \hspace{2cm}&\\\\ u_{3} &=90 \\\\ a+2 d &=90\ldots(1) \\\\ u_{5} &=80 \\\\ a+4 d &=80\ldots(2) \\\\ (2)-(1), & \\\\ 2 d &=-10 \\\\ \therefore\ d &=-5\\\\ \text { Substituting } d &=-5 \text { in equation }(1), \\\\ a-10 &=90 \\\\ \therefore a &=100 \\\\ \text{ (b) }\hspace{3.5 cm} S_m &=S_{m+1} \\\\ m\{200+(m-1)(-5)\} &=(m+1)\{200+m(-5)\} \\\\ m\{200+(m-1)(-5)\} &=(m+1)\{200+m(-5)\} \\\\ m\{200-5 m+5\} &=(m+1)\{200-5 m\} \\\\ m(205-5 m) &=(m+1)(200-5 m) \\\\ 205 m-5 m^{2} &=200 m-5 m^{2}+200-5 m \\\\ m &=20 \end{aligned}$



11. An arithmetic progression contains $25$ terms and the first term is $-15$. The sum of all the terms in the progression is $525$. Calculate
    (a) the common difference of the progression,
    (b) the last term in the progression,
    (c) the sum of all the positive terms in the progression.





$\begin{aligned} \text{(a) In an A.P.,}&\\\\ a &=-15 \\\\ a+12 d &=21 \\\\ S_{25} &=525 \\\\ \dfrac{25}{2}\{2 a+2 y d\} &=525 \\\\ -15+12 d &=21 \\\\ 12 d &=36 \\\\ d &=3 \text{(b) Last term} =u_{25}\\\\ &=a+2 y d \\\\ &=-15+72 \\\\ &=57 \end{aligned}$ Let the last non-positive term be $u_{n}$. $\begin{aligned} \therefore u_{n} & \le 0 \\\\ -15+(n-1) 3& \le0 \\\\ (n-1) 3 &\le15 \\\\ n-1 & \le 5\\\\ n & \le6 \\\\ \therefore\ u_{6} & \text{ is the last non-positive term.}\\\\ \therefore\ u_{7} &=u_{6}+3=3\\\\ \therefore\ u_{7} & \text{ is the first positive term.}\\\\ \therefore\ \text { required sum } &=u_{7}+u_{8}+\ldots+u_{25} \\\\ &=\dfrac{19}{2}\{3+57\} \\\\ &=570 \end{aligned}$



12. If the sum of $n$ terms of an A.P. is $n P+\dfrac{1}{2} n(n-1) Q$, where $P$ and $Q$ are constants, find the common difference.





$\begin{aligned} \text{In an A.P.,}&\\\\ S_{n} &=n P+\dfrac{1}{2} n(n-1) 2 \\\\ S_{1} &=P+\dfrac{1}{2}(1-1) Q=P \\\\ S_{2} &=2 P+\dfrac{1}{2}(2)(2-1) Q \\\\ &=2 P+Q \end{aligned}$ Let the common difference be $d$. $\begin{aligned} \therefore d &=u_{2}-u_{1} \\\\ &=S_{2}-S_{1}-S_{1} \\\\ &=S_{2}-2 S_{1} \\\\ &=2 P+Q - 2 P \\\\ &=Q \end{aligned}$



13. If the sum of first $p$ terms of an A.P. is equal to the sum of the first $q$ terms where $p \neq q$, then find the sum of the first $(p+q)$ terms.





$\begin{aligned} S_p&=S_q, \quad (p\ne q)\\\\ \dfrac{p}{2}\left\{2 a+(p-1) d\right\} &=\dfrac{q}{2}\left\{2 a+(q-1) d\right\} \\\\ p\left\{2 a+(p-1) d\right\} &=q\left\{2 a+(q-1) d\right\} \\\\ 2 a p+(p-1) p d &=2 a g+(q-1) q d \\\\ \left(p^{2}-p-q^{2}+q\right) d &=2 a q-2 a p \\\\ \left(p^{2}-q^{2}-p+q\right) d &=-2 a(p-q) \\\\ \left[(p-q)(p+q)-(p-q)\right] d &=-2 a(p-q)\\\\ (p-q)(p+q-1) d&=-2 a(p-q) \\\\ \quad \text { Since } p\ne q,\ & p-q=0. \\\\ \therefore\ (p+q-1) d&=-2 a \\\\ \therefore\ 2 a+(p+q-1) d&=0\\\\ \therefore\ S_{p+q}=\dfrac{p+q}{2}\{2 a +(p+q&-1) d\}=0 \\\\ \end{aligned}$



14. Prove that the sum of $n$ arithmetic means between two numbers is $n$ times the single A.M. between them.





Let $a$ and $b$ be two given numbers. Let $x_{1}, x_{2}, x_{3}, \ldots x_{n}$, be A.M.s between a and b. $\therefore \ a, x_{1}, x_{2}, x_{3}, \ldots x_{n}, b$ are in A.P. $\begin{aligned} a+x_{1}+x_{2}+x_{3}+\ldots+x_{n}+b&=\dfrac{n+2}{2}(a+b) \\\\ x_{1}+x_{2}+x_{3}+\ldots+x_{n} &=\dfrac{(n+2)(a+b)}{2}-(a+b)\\\\ &=(a+b)\left[\dfrac{n+2}{2}-1\right] \\\\ &=(a+b)\left[\dfrac{n+2-2}{2}\right]\\\\ &=\dfrac{n}{2}(a+b)\\\\ &=n\left(\dfrac{a+b}{2}\right)\\\\ &=n \text{ times the single A.M. between } a \text{ and } b \end{aligned}$



15. The first term of an A.P. is $x$, the second term is $y$ and the last term is $z$. Show that the sum of the A.P. is $\dfrac{(y+z-2 x)(z+x)}{2(y-x)}$.





Let the comnon difference of given A.P. be $d$. $\begin{aligned} &u_{1}=x \\\\ &u_{2}=x+d \\\\ &y=x+d \\\\ &d=y-x \\\\ &u_{n}=x+(n-1) d \\\\ &z=x+(n-1) d \\\\ &n-1=\dfrac{z-x}{d} \\\\ &z-x\\\\ n &=\dfrac{z-x}{y-x}+1 \\\\ &=\dfrac{z-x+y-x}{y-x} \\\\ &=\dfrac{y+z-2 x}{y-x} \\\\ S_{n} &=\dfrac{n}{2}(x+z) \\\\ &=\dfrac{(y+z-2 x)(x+z)}{2(y-x)} \end{aligned}$



16. The ratio of the sums of $m$ and $n$ terms of an A.P. is $m^{2}: n^{2}$. Show that the ratio of $m^{\text {th }}$ and $n^{\text {th }}$ terms is $(2 m-1):(2 n-1)$.





In an A.P., $\dfrac{S_{m}}{S_{n}}=\dfrac{m^{2}}{n^{2}}$ Let $S_{m}=k m^{2}$ and $S_{n}=k n^{2}$. Let $S_{m}=k m^{2}$ and $S_{n}=$ where $k$ is a constant. $\begin{aligned} u_{m} &=S_{m}-S_{m-1} \\\\ &=k m^{2}-k(m-1)^{2} \\\\ &=k m^{2}-k\left(m^{2}-2 m+1\right) \\\\ &=k m^{2}-k m^{2}+2 k m+k \\\\ &=k(2 m-1)\\\\ u_{n} &=S_{n}-S_{n-1} \\\\ &=k n^{2}-k(n-1)^{2} \\\\ &=k n^{2}-k\left(n^{2}-2 n+1\right) \\\\ &=k n^{2}-k n^{2}+2 k n-k \\\\ &=k(2 n-1) \\\\ \therefore\ \dfrac{u_{m}}{u_{n}} &=\dfrac{k(2 m-1)}{k(2 n-1)} \\\\ &=\dfrac{2 m-1}{2 n-1} \end{aligned}$



17. The sum of three numbers in an A.P. is $12$ and the sum of their cubes is $408$. Find the numbers.





Let the three numbers be $x, y$ and $z$. $\therefore x, y, z$, are in $A P$. Let the common difference be $d$. $\begin{aligned} \therefore\ y =x+d \text { and } z&=x+2 d \\ x+y+z &=12 \\ 3 x+3 d &=12 \\ x+d &=4 \\ \therefore\ x &=4-d\\ x^{3}+y^{3}+z^{3} &=408 \\ x^{3}+(x+d)^{3}+(x+2 d)^{3} &=408 \\ (4-d)^{3}+4^{3}+(4-d+2 d)^{3} &=408 \\ (4-d)^{3}+(4+d)^{3} &=344 \\ 64-48 d+12 d^{2}-d^{3}+64+48 d+12 d^{2}+d^{3} &=344 \\ 24 d^{2} &=216 \\ d^{2} &=9 \\ \therefore\ d &=\pm 3 \end{aligned}$ When $d=-3,\ x= 4-(-3) = 7$. Hence, the numbers are $7,4$ and $1$. When $d=3,\ x= 4-(3) = 1$. Hence, the numbers are $1, 4$ and $7$.