# Arithmetic progression class 10 word problems

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 $a_n = 5 - 6n$

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.

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?

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

Sum of nth terms
$S_n = \frac {n}{2} [2a + (n � 1) d]$
$S_{20} = \frac {20}{2} [2(a) + (20 � 1) d]$

$S_{20} = 1150$

Question 4.
If the sum of 7 terms of an A. P. is 49 and that of 17 terms is 289, find the sum of n terms.

Sum of nth terms
$S_n = \frac {n}{2} [2a + (n � 1) d]$

$S_7 = \frac {7}{2} [2a + 6d]=49$
or
$2a + 6d=14$--(1)

$S_{17} = \frac {17}{2} [2a + 16d]=289$
or
$2a + 16d=34$ ---(2)
Solving (1) and (2)
a=1 ,d=2
Now
Sum of nth terms
$S_n = \frac {n}{2} [2a + (n � 1) d]$

$=\frac {n}{2} [2 + (n � 1) 2]=n^2$

Question 5.
In an A. P., the sum of first n terms is $\frac {3n^2}{2} + \frac {13n}{2}$. Find its 25th term.

$T_{25} = S_{25} - S_{24}$ $=\frac {3(25)^2}{2} + \frac {13 \times 25}{2} - \frac {3(24)^2}{2} + \frac {13 \times 24}{2}=80$

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.

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.

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.

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=?,$S_n=5310$
$S_n = \frac {n}{2} [2a + (n � 1) d]$
$5310=\frac {n}{2} [354 + (n-1)-3]$
$3n^2 -357n +10620 = 0$
$n=\frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
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.

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.

a=8
$a_n=33$
$S_n=123$
Now
$S_n = \frac {n}{2} [2a + (n � 1) d]$

$S_n=\frac {n}{2} [a + a_n]$
$123= \frac {n}{2} [8 + 33]$
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?

a=17
$a_n=350$
d=9
$a_n= a + (n-1)d$
$350 = 17 + (n-1)9$
n=38
Now
$S_n=\frac {n}{2} [a + a_n]$
$=\frac {38}{2} [17 + 350]=6973$

Question 12.
In an A. P., the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A. P.

a=2
$a_n=29$
$S_n=155$
Now
$S_n=\frac {n}{2} [a + a_n]$
$155= \frac {n}{2} [2+ 29]$
n=10

Now
$a_n= a + (n-1)d$
$29=2 + (10-1)d$
d=3

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.

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.

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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### Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20