Arithmetic Progressions Chapter 5 Example
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Example 1 : For the AP : 3/2 , 1/2 , – 1/2 , – 3/2 , . . ., write the first term a and the common difference d.
solution:- first term of Ap=a=3/2]
Common Difference=3/2 - 1/2
d=(3-1)/2
d=2/2
d=1
Example 2 : Which of the following list of numbers form an AP? If they form an AP,
write the next two terms :
(i) 4, 10, 16, 22, . . . (ii) 1, – 1, – 3, – 5, . . .
(iii) – 2, 2, – 2, 2, – 2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . .
solution:-
(i) 4, 10, 16, 22, . . .
d1=10-4=6
d2=16-10=6
d3=22-16=6
d1=d2=d3
Hence total common difference is Equal then it is in AP.
First next term=previous term +common difference
=22+6=28
second next term=previous term +common difference
=28+6=34
(ii) 1, – 1, – 3, – 5, . . .
d1=-1-1=-2
d2=-3-(-1)=-3+1=-2
d3=-5-(-3)=-5+3=-2
d1=d2=d3
Hence total common difference is Equal then it is in AP.
First next term=previous term +common difference
=-5+(-2)=-5-2=-7
second next term=previous term +common difference
=-7+(-2)=-7-2=-9
(iii) – 2, 2, – 2, 2, – 2, . . .
d1=2-(-2)=2+2=4
d2=-2-2=-4
d1≠d2
Hence total common difference is not Equal then it is not in AP.
term is not find out because seqence is not in AP.
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . .
d1=1-1=0
d2=1-1=0
d3=2-1=1
d1=d2 ≠ d3
Hence total common difference is not Equal then it is not in AP.
term is not find out because seqence is not in AP.
Example 3 : Find the 10th term of the AP : 2, 7, 12, . .
solution:- Given that a=2 and n=10
d=7-2 =5
we know that
An=a+(n-1)d
A10= 2+(10 -1)×5
=2+9×5
A10=45
Example 4 : Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer.
solution:- Given that a=21 and n=?
d=18-21 =-3, An=0
we know that
An=a+(n-1)d
0=21+(n-1)(-3)
0=21-3n+3
0=24-3n
3n=24
n=24/3
=8 ans
Example 5 : Determine the AP whose 3rd term is 5 and the 7th term is 9.
solution:-
Given that
3rd term of the AP=5
means,a3=5
a+2d=5 ............(i)
and 7th term of the AP=9
means,a7=9
a+6d= 9 ..........(ii)
from Equation (i) and (ii)
a+2d=5 ............(i)
a+6d= 9 ..........(ii)
After solving
d=1 ,a=3
AP is
a1=a=3
a2=a+d=3+1=4
a3=a2+2d=3+2×1=5
a4=a+3d=3+3×1=6
a5= a+4d=3+4×1=7
so on ........
Hence, the required AP is 3,4,5,6,7.............................. Ans.
Example 6 : Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . .
solution:-
Given that
the term of the Ap is 5,11,17,23........
a=5,
d=11-5=6
An=301 then n=?
An=301
An=a+(n-1)d
301 =5+(n-1)×6
301 =5+6n-6
301=-1+6n
6n=302
n=302/6
But n should be a positive integer . So, 301 is not a term of the given list of
numbers. Ans
Example 7 : How many two-digit numbers are divisible by 3?
solution
let the two digit number which is divisible by 3 is
12,15,18,21................99.
a=12,
d=15-12=3
An=99 then n=?
An=99
An=a+(n-1)d
99=12+(n-1)×3
99=12+3n-3
99=9+3n
3n=99-9
3n=90
n=30 ans....
So, there are 30 two-digit numbers divisible by 3.
Example 8 : Find the 11th term from the last term (towards the first term) of the AP : 10, 7, 4, . . ., – 62.
solution:-
Given that
the term of the Ap is 10,7,4.........,-62.
for finding 11th term 0f Ap so reverse this Ap
-62,............ 4,7,10.
from this sequence
a=-62,
d=an-a(n-1)=10-7=3
a11=a+10d
=-62+10×3
=-62+30
=-32
*Alternative Solution
Comming soon
Example 9 : A sum of RS 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.
Solution:-
Comming Soon
Example 10 : In a flower bed, there are 23 rose plants in the first row, 21 in the
second, 19 in the third, and so on. There are 5 rose plants in the last row. How many
rows are there in the flower bed?
Comming soon...
Arithmetic Progressions Math Exercise-5.1( Click here)Solution
