## Tuesday, April 23, 2013

### 53-03 Basics of Arithmetic & Geometric Progression (Grades 9 to 12) Part-03

Blog-53
Dear Students,

In the Previous Blog we had seen some important concepts of Arithmetic & Geometric Progression.
1) Click Here to Read the Blog-60 (

E) Find tn for an AP where t5  = 19 & t15  = -21.

Solution:

1) Let " a " be the first term & " d " be the common difference.
2) So, t= a + ( n - 1 ) d
3) According to the problem,
t5   = a +  ( 4 )  d =  19
t15 = a + (14 )  d = -21
(-)  (-)               (+)
--------------------------------------------
- 10  d  =  40
-  d  =  4
d  =  - 4
4) We know that
t5   =  a +  ( 4 ) d =  19
a + 4 * (- 4) =  19
a  - 16 =  19
a   =  19 + 16
a   =  35
5) So,  t= a + ( n - 1 ) d
t= 35 + ( n - 1 ) * (- 4)
t= 35 - 4 n + 4
t= 39 - 4 n
6) Answer: Here the nth term of an AP is  tn =  39 - 4 n

F) Find, how many 3 digit natural numbers are divisible by 3.

Solution:

1) We know that the lowest 3 digit number is 100 & the largest one is 999.
2) So the numbers divisible by 3 between 100 & 999 are
102, 105, 108, ----, 996, 999.
3) Here a = 102, d = 3 and last term = 999.
4) So, from  t= a + ( n - 1 ) d, we have,
999 = 102 + ( n - 1 ) * 3
3 *  ( n - 1 )  =  999 - 102
3 *  ( n - 1 )  =  897
( n - 1 )  =  897 / 3
( n - 1 )  =  299
n    =  299 + 1
n    =  300
5) Answer: Here there are 300 three digit numbers which are divisible by 3.

Problems on Summation will be published in the next Blog.

Anil Satpute