Question

Solve the following questions

1. 11, 8, 5, 2, . . . In this A.P. which term is number -151?

2. Find how many three-digit natural numbers are divisible by 5.

3. The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41th

term of that A.P.

4. In the natural numbers from 10 to 250, how many are divisible by 4?

5. In an A.P. 17th term is 7 more than its 10th term. Find the common difference.​

Answer 1

1. Given AP=11,8,5,2......
D= a2-a1=8-11=-3,
An=[a+(n-1)d]
-151=[11+(n-1)-3]
-151=11-3n+3
-151=14-3n
-151-14=-3n
-165=-3n
3n=165
n=165/3
n=55
Hence, -151 is 55th term of the given AP
2.AP=100,105,110......995
D=a2-A1=105-100=5
An=[a+(n-1)d]
995=[100+(n-1)5]
995=[100+5n-5]
995=[95+5n]
995-95=5n
900=5n
5n=900
n=900/5
n=180
Hence there are 180 three-digit numbers that are divisible by 5.
3. Given,
11th term=a+10d=16,—>1
21st term=a+20d=29,—>2
Subtract equation 1 from 2,
a+20d-(a+10d)=29-16
a-a+20d-10d=13
10d=13
From equation 1,
a+10d=16
a+13=16
a=16-13
a=3
41st term=a+40d=a+4(10d)=3+4(13)=3+52=55
Hence, the 41st term of the given AP =55.
4. AP=12,16,20......248
D=4,
An=[a+(n-1)d]
248=[12+(n-1)4]
248=[12+4n-4]
248=8+4n
248-8=4n
4n=240
n=240/4
n=60
Hence there are 60 natural numbers from 10 to 250 that are divisible by 4.
5.10th term=a+9d=x—>equation 1
17th term=a+16d=x+7–>equation 2
Subtracting equation 1 from 2,
A+16d=x+7
A+9d =x
—————
7d=7
d=7/7
d=1.
Hence common difference(d)=1.