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In the Previous Blog, we had seen some important concepts of Arithmetic & Geometric Progression.E) Find tn for an AP where t5 = 19 & t15 = -21.
Solution:
1) Let " a " be the first term & " d " be a common difference.
2) So, tn = 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, tn = a + ( n - 1 ) d
tn = 35 + ( n - 1 ) * (- 4)
tn = 35 - 4 n + 4
tn = 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 tn = 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 that are divisible by 3.
In the next part, we will see a few examples and some essential formulae.
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