add three,four,five and six consecutive even numbers to get a pattern
Answers
Answer:
can you give the question proper way,
it make people to answer easily
Answer:
They form a pattern of [(n)a + n(n - 1)] where 'n' is the number of terms and 'a' is the first term.
Step-by-step explanation:
We can solve this Question in many ways, I will show you my Favorite method, that is, Arithematic Progression(A.P.)
If you don't know AP, let me brief it for you
An AP is a set of numbers having a common difference between them
Ex:- 1, 2, 3, 4,......
Here a (known as first term) = 1
Common difference d = 2 - 1 = 1
In case of common difference it will be the same for all the numbers
Another example,
1, 3, 5, 7
a = 1
d = 3 - 1 = 2 or 7 - 5 = 2 or 5 - 3 = 2
Only thing is, they have to be consecutive
Just remember that their Sum will be
(n/2)[2a + (n - 1)d]
where 'a' is the 1st term, 'n' is the number of terms and 'd' is the common difference.
Let the first term be 'a' which is an even number
we know that the difference between two consecutive even numbers is 2
Ex:- 4 - 2 = 2 or 6 - 4 = 2
so, d = 2
Now,
We have 4 cases
Case 1
1st term = a
No. of terms (n) = 3
Common difference (d) = 2
we know that,
S(nth) = (n/2)[2a + (n - 1)d]
S(3rd) = (3/2)[2a + (3 - 1)2]
S(3rd) = (3/2)[2a + (2)2]
S(3rd) = (3/2)[2a + 4]
S(3rd) = (3/2)[2(a + 2)]
S(3rd) = (3)[a + 2]
S(3rd) = (3a + 6)
Case 2
1st term = a
No. of terms (n) = 4
Common difference (d) = 2
we know that,
S(nth) = (n/2)[2a + (n - 1)d]
S(4th) = (4/2)[2a + (4 - 1)2]
S(4th) = (2)[2a + (3)2]
S(4th) = (2)[2a + 6]
S(4th) = (4a + 12)
Case 3
1st term = a
No. of terms (n) = 5
Common difference (d) = 2
we know that,
S(nth) = (n/2)[2a + (n - 1)d]
S(5th) = (5/2)[2a + (5 - 1)2]
S(5th) = (5/2)[2a + (4)2]
S(5th) = (5/2)[2a + 8]
S(5th) = (5/2)[2(a + 4)]
S(5th) = (5)[a + 4]
S(5th) = (5a + 20)
Case 4
1st term = a
No. of terms (n) = 6
Common difference (d) = 2
we know that,
S(nth) = (n/2)[2a + (n - 1)d]
S(6th) = (6/2)[2a + (6 - 1)2]
S(6th) = (3)[2a + (5)2]
S(6th) = (3)[2a + 10]
S(6th) = (6a + 30)
So, now we get
S(3rd) = (3a + 6)
S(4th) = (4a + 12)
S(5th) = (5a + 20)
S(6th) = (6a + 30)
Now, we see a pattern
We see that, as the sum of numbers of consecutive even numbers increase, they are of the form [(n)a + n(n - 1)]
That is, if I want the sum of 2 consecutive numbers,
then it will be,
S(2nd) = (2)a + 2(2 - 1)
S(2nd) = 2a + 2(1)
S(2nd) = 2a + 2
Let's check that say a = 2
Then Sum of 2 consecutive numbers = 2 + 4 = 6
And according to our equation,
S(2nd) = 2(2) + 2 = 4 + 2 = 6
So, They are of the form [(n)a + n(n - 1)]
It could be a bit hard try to read it again and you will surely get it
Hope it helped and you understood it........All the best