Exercise 5.3 Practice
Sum of n Terms ($S_n$) Practice
Q1
Find the sum of the following arithmetic progressions:
(i) 4, 9, 14, … to 15 terms.
$a=4,\ d=5,\ n=15$
$S_n=\frac{n}{2}[2a+(n-1)d]$
$S_{15}=\frac{15}{2}[8+70]=585$
Sum = 585
(ii) −25, −20, −15, … to 18 terms.
$a=-25,\ d=5,\ n=18$
$S_{18}=\frac{18}{2}[-50+85]=315$
Sum = 315
(iii) 1.2, 2.0, 2.8, … to 40 terms.
$a=1.2,\ d=0.8,\ n=40$
$S_{40}=20[2.4+31.2]=672$
Sum = 672
(iv) $\frac1{12},\frac1{9},\frac1{6},\ldots$ to 10 terms.
$a=\frac1{12},\ d=\frac1{36},\ n=10$
$S_{10}=5\left(\frac16+\frac14\right)=\frac{25}{12}$
Sum = 25/12
Q2
Find the sums given below:
(i) 6 + 10 + 14 + … + 90
$a=6,\ d=4,\ l=90$
$90=6+(n-1)4 \Rightarrow n=22$
$S_{22}=11(96)=1056$
Sum = 1056
(ii) 40 + 37 + 34 + … + 7
$a=40,\ d=-3,\ l=7$
$7=40+(n-1)(-3) \Rightarrow n=12$
$S_{12}=6(47)=282$
Sum = 282
(iii) −6 + (−10) + (−14) + … + (−206)
$a=-6,\ d=-4,\ l=-206$
$-206=-6+(n-1)(-4) \Rightarrow n=51$
$S_{51}=51(-106)=-5406$
Sum = −5406
Q3
In an arithmetic progression:
(i) Given that the first term is 6 and the common difference is 4, find the number of terms if the nth term is 70. Also find the sum of these terms.
$70=6+(n-1)4 \Rightarrow n=17$
$S_{17}=\frac{17}{2}(6+70)=646$
n = 17, Sum = 646
(ii) Given that the first term is 9 and the 15th term is 57, find the common difference and the sum of the first 15 terms.
$57=9+14d \Rightarrow d=\frac{24}{7}$
$S_{15}=\frac{15}{2}(9+57)=495$
d = 24/7, Sum = 495
(iii) Given that the 14th term of an arithmetic progression is 50 and the common difference is 3, find the first term and the sum of the first 14 terms.
$50=a+13×3 \Rightarrow a=11$
$S_{14}=7(61)=427$
a = 11, Sum = 427
(iv) Given that the third term of an arithmetic progression is 18 and the sum of the first 12 terms is 216, find the common difference and the 12th term.
$a+2d=18$
$216=6(2a+11d)$
$d=2,\ a=14$
$a_{12}=36$
(v) Given that the common difference is 6 and the sum of the first 10 terms is 210, find the first term and the 10th term.
$210=5(2a+54)$
$a=-6$
$a_{10}=48$
(vi) Given that the first term is 3, the common difference is 7 and the sum of n terms is 175, find n and the nth term.
$175=\frac{n}{2}[6+(n-1)7]$
$n=10$
$a_{10}=66$
(vii) Given that the first term is 10, the nth term is 85 and the sum of n terms is 285, find n and the common difference.
$285=\frac{n}{2}(95) \Rightarrow n=6$
$d=15$
(viii) Given that the nth term is 6, the common difference is 3 and the sum of n terms is −36, find n and the first term.
$n=6,\ a=-9$
First term = −9
(ix) Given that the first term is 5, the number of terms is 12 and the sum of all the terms is 390, find the common difference.
$390=6[10+11d]$
$d=5$
(x) Given that the last term is 40, the sum of all the terms is 220 and there are 10 terms, find the first term.
$220=5(a+40)$
$a=4$
Q4
How many terms of the arithmetic progression 12, 20, 28, … must be taken so that their sum is 800?
$800=\frac{n}{2}[24+(n-1)8]$
Number of terms = 10
Q5
The first term of an arithmetic progression is 8, the last term is 68 and the sum of all the terms is 608. Find the number of terms and the common difference.
$608=\frac{n}{2}(76) \Rightarrow n=16$
$d=4$
Q6
The first and the last terms of an arithmetic progression are 25 and 400 respectively. If the common difference is 15, find the number of terms and the sum of the progression.
$400=25+(n-1)15 \Rightarrow n=26$
Sum = 5525
Q7
Find the sum of the first 18 terms of an arithmetic progression whose common difference is 9 and whose 18th term is 170.
$a=170-17×9=17$
Sum = 1683
Q8
Find the sum of the first 60 terms of an arithmetic progression whose second and third terms are 11 and 15 respectively.
$d=4,\ a=7$
Sum = 7500
Q9
If the sum of the first 6 terms of an arithmetic progression is 36 and that of the first 18 terms is 324, find the sum of the first n terms.
$a=1,\ d=2$
$S_n=n^2$
Q10
Show that the sequence defined by $a_n = 5 + 6n$ forms an arithmetic progression. Also find the sum of the first 12 terms.
$a_1=11,\ a_2=17$
Sum = 528
Q11
If the sum of the first n terms of an arithmetic progression is given by
\( S_n = 6n - n^2 \),
find the first term, the second term and the nth term of the progression.
$S_1 = 6(1)-1^2 = 5 \Rightarrow a_1 = 5$
$S_2 = 12-4 = 8$
$a_2 = S_2 - S_1 = 3$
$a_n = S_n - S_{n-1} = 7 - 2n$
First term = 5, Second term = 3, nth term = 7 − 2n
Q12
Find the sum of the first 50 positive integers which are divisible by 4.
Required AP: 4, 8, 12, … (50 terms)
$a=4,\ d=4,\ n=50$
$S_{50}=\frac{50}{2}[8+49(4)]$
Sum = 5100
Q13
Find the sum of the first 20 multiples of 9.
AP: 9, 18, 27, …
$a=9,\ d=9,\ n=20$
$S_{20}=\frac{20}{2}[18+19(9)]$
Sum = 1890
Q14
Find the sum of all even natural numbers between 10 and 100.
Even numbers: 12, 14, …, 98
$a=12,\ l=98,\ d=2$
$n=\frac{98-12}{2}+1 = 44$
Sum = 2420
Q15
A contractor is fined ₹150 on the first day, ₹200 on the second day, ₹250 on the third day and so on.
Find the total penalty that he has to pay if the delay lasts for 25 days.
$a=150,\ d=50,\ n=25$
$S_{25}=\frac{25}{2}[300+24(50)]$
Total penalty = ₹18,750
Q16
₹900 is to be distributed among 9 students in such a way that each student gets ₹30 less than the previous one.
Find the amount of money received by each student.
$S_9 = 900,\ d = -30$
$900=\frac{9}{2}[2a-240]$
$a = 220$
Amounts: 220, 190, 160, 130, 100, 70, 40, 10, −20
Q17
There are 4 sections in each class from Class I to Class X. Each section plants trees equal to its class number.
Find the total number of trees planted.
Trees per class form the AP: 4, 8, 12, …, 40
$n=10$
$S=\frac{10}{2}(4+40)$
Total trees = 220
Q18
A spiral is made up of 15 semicircles. The radii of the semicircles are 1 cm, 2 cm, 3 cm, … .
Find the total length of the spiral.
(Take \( \pi = \frac{22}{7} \))
Length of each semicircle = πr
AP: π, 2π, 3π, … (15 terms)
$S_{15}=\frac{15}{2}(π+15π)=120π$
Total length = 377.14 cm
Q19
300 logs are stacked in such a way that the bottom row contains 25 logs, the next row 24 logs and so on.
Find the number of rows and the number of logs in the top row.
$a=25,\ d=-1,\ S_n=300$
$300=\frac{n}{2}(51-n)$
$n=15$
Rows = 15, Logs in top row = 11
Q20
In a potato race, the first potato is placed 4 m from the starting point and the remaining potatoes are placed 2 m apart.
There are 12 potatoes. Find the total distance covered by the competitor to pick up all the potatoes and return to the starting point.
Distances one way: 4, 6, 8, … (12 terms)
$S_{12}=\frac{12}{2}[8+11(2)]=180$
Total distance = $2×180$
Total distance = 360 m