Ncert – Class 10 -Arithmetic Progressions-Hots Question

Question 1

In a theater, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?

Solution

Number of seats in 1st row = 20
Number of seats in 2nd row = 20 + 2 = 22
Number of seats in 3rd row = 22 + 2 = 24
Total number of rows in theatre (n) = 30
a = 20, d = 2 and n = 30
number of seats in last row be “l”
n = [(l-a)/d] + 1
30 = [(l-20)/2] + 1
(30-1) = (l-20)/2
29 (2) = l-20
l-20 = 58
l = 58 + 20
l = 78
So, the number of seats in the last row is 78.

Question 2

If the first, second and last terms of the AP are 5, 9, 101, respectively, find the
total number of terms in the AP.

Solution

Given: First term, a = 5
Common difference, d = 9 – 5 = 4
Last term, an = 101
Now, we have to find the value of “n”.
Hence, an = a+(n -1)d
Substituting the values, we get
5+(n -1)4 = 101
5+4n-4 = 101
4n+1= 101
4n = 100
n=100/4 = 25
Hence, the number of terms in the AP is 25

Question 3

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

a=51 ,d=2.

Question 4

A man saved 16500 in ten years. In each year after the first he saved 100 more than he did in the
preceding year . How much did he save in the first year ?

Solution

16500==5(2a+900)
16500= 10a+4500
10a=12000
a= 1200
Hence he saves RS. 1200 in first year

Question 5

The sum of three numbers in an A.P is 24 and the sum of their cubes is 1968.find the numbers.

Solution

LET THREE NUMBER IN A.P. ARE
a-d, a, a+d
sum=(a-d)+a+(a+d)=24
3a=24
a=8
SUM OF THEIR CUBES = 1968
(a-d)³+a³+(a+d)³=1968
3a³+6ad²=1968
3×8³+6x8xd²=1968
1536+48d²=1968
48d²=432
d²=9, d = ±3
HENCE THE REQUIRED NUMBERS A
a-d, a, a+d
8-3, 8, 8+3
5, 8, 11 OR 11,8,5

Question 6

The longest side of a right triangle is 5 more than 4 times the shortest side. The third side is 10 less than the sum of the other two. Let the length of the shortest side be s cms. Find the three sides of triangle.

Solution

S, 4s+5,s+4s+5-10; using PGT (4s+5)² = s² + (5s-10)² ; s(s-9)=0; s=9 cm
,hypo=41 cm, other leg=40 cm

Question 7

Is it possible to design a rectangular park whose length is thrice its breadth and area is 972 m ²? If yes, find length and breadth

Solution

L X B = 972; 3B2 – 972 = 0; b = 18 m, l=54 m

Question 8

The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.

Solution

1st term = a – d = 9 – 7 = 2
2nd term = a = 9
3rd term = a + d = 9 + 7 = 16
So, the first three terms are 2, 9, 16

Question 9

The ratio of 6th and 8th term of an A.P. is 7:9. Find the ratio of 9th term to 13th term.

Solution

ratio of 9th term to 13th term
𝑡9
: 𝑡13 = (a + 8d) / (a + 12d)
= (2d + 8d) / (2d + 12d)
= 10d / 14d
= 5 / 7
= 5 : 7
so, the ratio of 9th term to 13th term is 5 : 7

Question 10

Prove that 7, 11, 15, 19, 23 is an AP.

Solution

Given sequence: 7, 11, 15, 19, 23.
To prove that the sequence is AP, find the common difference between
two consecutive terms.
d = 11 – 7 = 4
d = 15 – 11 = 4
d = 19 – 15 = 4
d = 23 – 19 = 4
Hence, 7, 11, 15, 19, 23 is an AP with a common difference of 4