Скачать или смотреть Class 10 Maths Exercise 5.2 Q18,19,20 Arithmetic Progressions part 18 NCERT

In this video, we will be solving questions of exercise 5.2 Arithmetic Progressions.
For any questions or doubts, feel free to ask in the comments below.


Question 18:
The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th
and 10th terms is 44. Find the first three terms of the A.P.
Solution 18:
Let a be the first term and d the common difference.
Given, a4 + a8 = 24
(a + 3d) + (a + 7d) = 24
⇒2a + 10 d = 24
⇒a+5d =12.................Equation (1)
Also, a6 + a10 = 44
(a + 5d) + (a + 9d) = 44
⇒2a + 14d = 44
⇒a + 7d = 22 .................Equation (2)
On subtracting Equation (1) from (2), we obtain
a + 5 d –(a+7d) = 12− 22
a−a +5d −7d = −10
−2d = −10
2d = 10
d = 5
By Substituting the value of d=5 in Equation (1), we obtain
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
The first three terms are a, (a + d) and (a + 2d)
Substituting the values of a and d, we get – 13, (– 13 + 5) and (– 13 + 2 × 5)
i.e., – 13, – 8 and – 3
Therefore, the first three terms of this A.P. are −13, −8, and −3.
Question 19:
Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?


Solution 19:
From the Given Data, incomes received by Subba Rao in the years 1995,1996,1997… are 5000, 5200, 5400, ……… 7,000
From Observation, Common difference, d= 200
a = 5000
d = 200
Let after nth year, his salary be Rs 7000.
Hence an = Rs.7000
We know that the nth term of an A.P. Series,
an = a + (n − 1) d
By Substituting above values,
7000 = 5000 + (n − 1) 200
200(n − 1) = 7000−5000
200(n−1)=2000
(n − 1) = 2000/200 = 10
n = 11
Therefore, in 11th year, his salary will be Rs 7000.


Question 20:
Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75.
If in the nth week, her weekly savings become Rs 20.75, find n.
Solution 20:
From the given data, Ramkali’s savings in the consecutive weeks are
.5, .5 .1.75,
.5 2 .1.75, .5 3 .1.75...
Rs Rs Rs
Rs Rs Rs Rs

   
Hence nth Term is Rs. 5+(n−1) × Rs. 1.75 = Rs. 20.75
Now from the above we know that
a = 5
d = 1.75
an = 20.75
n = ?
We know that the nth term of an A.P. Series,
 
 
 
 
n a a n 1 d
20.75 5 n 1 1.75
15.75 n 1 1.75
15.75 1575
n 1
1.75 175
63
9
7
  
  
 
  
 
n − 1 = 9
n = 10
Hence, n is 10.