Given below are the Class 10 Maths word Problems for Arithmetic Progression Question 1. Find the sum of m terms of an A. P. whose nth terms is given by $a_n = 5 - 6n$ Solution
Given
$a_n = 5 - 6n$
$a_1= 5-6=-1$
$a_2= 5-12=-7$
Common difference = $a_2 - a_1= -7 +1 = -6$
Now
$S_m= \frac {m}{2}[2a + (m-1)d]$
$S_m= \frac {m}{2} [-2 + (m-1)-6]= m (2 - 3m)$
Question 2. Find the sum of all even integers between 101 and 999. Solution
The series will be
102, 104,106,....998
Now a=102, d=2 ,n=?
Now from nth term formula
$a_n= a + (n-1)d$
$998 = 102 + (n-1)2$
$n=449$
Now
$S_n= \frac {n}{2}[2a + (n-1)d]$
$=\frac {449}{2}[204 + (449-1)2]=246950$
Question 3. In an A. P, if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 term? Solution
Given
$
a_5 = 30$, $a_12 = 65$
Now
$
a_5 = 30$
$ a + (5 - 1)d=30$
$a+ 4d=30$ -(1)
Also
$a_12 = 65$
$
a + (12 - 1)d = 65
$
$a + 11d = 65$ -(2)
Solving (1) and (2)
a=10, d=5
Question 6. A man arranges to pay off a debt of Rs 3600 by 40 annual installments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the first installment. Solution
Sum of nth terms
$S_n = \frac {n}{2} [2a + (n � 1) d]$
40 installment =3600
$S_{40} = \frac {40}{2} [2a + 39d]=3600$
or
$2a + 39d=180$--(1)
30 installment = 3600 - (1/3) x 3600 = 2400
$S_{30} = \frac {30}{2} [2a + 29d]=2400$
or
$2a + 29d=160$ ---(2)
Solving (1) and (2)
a=51 ,d=2
Question 7. There are 25 trees at equal distances of 5m in a line with a well, the distance of the well from the nearest tree being 10m. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees. Solution
For first tree =20 m distance is covered in to fro from well
For second tree =$2(10+ 1\times 5)=30$ m distance is covered in to fro from well
For third tree =$2(10+2 \times 5)=40$ m distance is covered in to fro from well
....
For 25th tree= $2(10 + 24 \times 5)=260$ m distance is covered in to-fro from well
So total distance
=20 + 30 + 40 +....+ 260
This is a A.P with a=20,d=10 and n=25
$S_n = \frac {n}{2} [2a + (n � 1) d]$
$S= \frac {25}{2}[40 + 24 \times 10]=3500$ m
Question 8. A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount. Solution
Amount counted in half an hour = Rs 180 * 30 = Rs.5400
Therfore the remaining to be calculated is = Rs10710 - Rs 5400 = Rs 5310
According to the question, after 30 min, counting rate start decreasing every minute by 3
So At 31st min ,counting rate is 177
At 32nd min, counting rate will be 174
At 33nd min, counting rate will be 171
This is clearly as A.P
177, 174,171.......
Here a=177, d=-3 ,n=?,$S_n=5310$
$S_n = \frac {n}{2} [2a + (n � 1) d]$
$5310=\frac {n}{2} [354 + (n-1)-3]$
$
3n^2 -357n +10620 = 0$
By using quadratic formula
$n=\frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
n= 59 or 60.
so we will use 59 because these are minutes
and itis the work is in 59 min.
So total time to count all the Rs= 59min + 30min = 1hr 29min
Question 9. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performances. If each price is Rs 20 less than its preceding prize, find the value of each prize. Solution
Let first prize is a,then seven prizes will be
a, a-20,a- 40,a-60,a-80,a-100,a-120
Now
$a+ a-20+a- 40+a-60+a-80+a-100+a-120=700$
$7a=1120$
$a= 160$
So prizes will be
160,140,120,100,80,60,40
Question 10. In an A. P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and d, the common differences. Solution
$S_n=\frac {n}{2} [a + a_n]$
$123= \frac {n}{2} [8 + 33]$
n=6
Also
$a_n= a + (n-1)d$
$33=8+ (6-1)d$
d=5
Question 11. The first and the last term of an A. P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? Solution
Question 13. In an A. P, the sum of first ten terms is -150 and the sum of its next ten terms are -550. Find the A. P. Solution
a = 6, d = -4
Question 14
The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7 : 15. Find the numbers. Solution
Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d)
According to the question.
$a-3d + a - d + a + d + a + 3d = 32
$
$4a = 32$
$a = 8$ ......(1)
Now,
$\frac {(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac {7}{15}$
$15(a^2 - 9d^2) = 7(a^2 - d^2)
$
$15a^2 - 135d^2 = 7a^2 - 7d^2$
$8a^2 = 128d^2$
Putting the value of a = 8 in above we get.
$
8(8)^2 = 128d^2$
$d^2 = 4$
$d=\pm 2$
Putting a=8, d=+2
the four consecutive numbers are
2, 6, 10 and 14
Putting a=8 ,d=-2 ,we get
the four consecutive numbers are
14, 10, 6 and 2
Summary
This Class 10 Maths Word Problems for Arithmetic Progression with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.
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