ARITHMETIC PROGRESSIONS
107
17. Find the 20th term from the last term of the AP:3,8,13,...,253
18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is
44. Find the first three terms of the AP.
19. Subba Rao started work in 1995 at an annual salary of 3 5000 and received an increment
of 200 each year. In which year did his income reach 7000?
20. Ramkali saved 5 in the first week of a year and then increased her weekly savings by
1.75. If in the nth week, her weekly savings become * 20.75, find n.
Answers
Answer:
ans is 25
Step-by-step explanation:
the sum of the 4 and 8 terms are in ap is 24
Answer:
17. a=3,d=5
253=3+5(n−1)
Or, 5(n−1)=250
Or, n−1=50
Or, n=51
So, the 20th term from the last term =51−19=32nd term
Now, n
32
=3+5×31=158
18.
a n = a + (n − 1) d
a 4 = a + (4 − 1) d
a 4 = a + 3d
Similarly,
a 8 = a + 7d
a 6 = a + 5d
a 10 = a + 9d
Sum of 4th and 8th term = 24 (Given)
a 4 + a 8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12.................... (i)
Sum of 6th and 10th term = 44 (Given)
a 6 + a 10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 ......................(ii)
From equation (i), we get,
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a 2 = a + d = − 13 + 5 = −8
a 3 = a 2 + d = − 8 + 5 = −3
19. It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are
5000,5200,5400,…
Here, a=5000
d=200
Let after n
th
year, his salary be Rs 7000.
Therefore, a
n
=a+(n−1)d
⇒7000=5000+(n−1)200
⇒200(n−1)=2000
⇒(n−1)=10
∴n=11
Therefore, in 11
th
year, his salary will be Rs 7000.
20. According to the question, a=5, d=1.75, a
n
=20.75
a
n
=a+(n−1)d
Subsituting the values in above equation,
20.75=5+(n−1)×1.75
15.75=(n−1)(1.75)
⇒n−1=
1.75
15.75
⇒n−1=9
⇒n=10
n=10.