Exercise 5.4 Practice
Sum of n Terms ($S_n$) Practice
Q1
Which term of the arithmetic progression 95, 90, 85, … will be the first term having a negative value?
$a=95,\ d=-5$
$a_n = 95+(n-1)(-5) < 0$
$100-5n<0 \Rightarrow n>20$
First negative term is the 21st term
Q2
The sum of the 4th and 10th terms of an arithmetic progression is 30 and their product is 176.
Find the sum of the first 20 terms of the arithmetic progression.
$a+3d + a+9d = 30 \Rightarrow a+6d=15$
$(a+3d)(a+9d)=176$
Solving gives $a=4,\ d= \frac{11}{6}$
$S_{20}=\frac{20}{2}[8+19(\frac{11}{6})]$
Sum = 410
Q3
A stair has steps which are 30 cm apart. The width of the steps decreases uniformly from
60 cm at the bottom to 36 cm at the top. If the total height of the stair is 3 m,
find the total length of wood used to make the steps.
Number of steps = $\frac{300}{30}+1 = 11$
Widths form AP: 60, 57.6, … , 36
$S_{11}=\frac{11}{2}(60+36)$
Total length of wood = 528 cm
Q4
The houses on one side of a street are numbered consecutively from 1 to 81.
Find the number of the house such that the sum of the numbers of the houses before it
is equal to the sum of the numbers of the houses after it.
Let the required house number be $x$
$S_{x-1} = S_{81} - S_x$
Solving gives $x=41$
Required house number = 41
Q5
A stadium has 20 rows of seats. The first row has 25 seats and each subsequent row has
3 seats more than the previous row. Find the total number of seats in the stadium.
$a=25,\ d=3,\ n=20$
$S_{20}=\frac{20}{2}[50+19(3)]$
Total seats = 1570
Q6
A pile of bricks is stacked in such a way that the bottom row has 30 bricks and each
row above it has one brick less than the row below it. If there are 18 rows,
find the total number of bricks in the pile.
$a=30,\ d=-1,\ n=18$
$S_{18}=\frac{18}{2}(30+13)$
Total bricks = 387
Q7
A student saves ₹50 in the first month, ₹75 in the second month and ₹100 in the third month
and continues saving in the same manner. Find the total savings after 18 months.
$a=50,\ d=25,\ n=18$
$S_{18}=\frac{18}{2}[100+17(25)]$
Total savings = ₹4,275
Q8
Find the sum of all numbers between 200 and 500 which are divisible by 7.
First number = 203, Last number = 497
$a=203,\ d=7$
$n=\frac{497-203}{7}+1=43$
Sum = 15050