Hiii, brainy users
A tricky question
without adding find the sum
1)1+3+5+7
2)1+3+5+7+9+11
3)1+3+5+7+9+11+13+15+17+19+21
CONTENT QUALITY ANSWER NEEDED
no spam answers from answers will be deleted on the spot
SamriddhaChandra:
7×7 = 14
it must be 7 +7 and not 7 * 7
my bro is doing these things sorry from him
I'm telling u the and
4×4 = 16
6 × 6 = 36
11 × 11= 121
these r the 3 answers. it is said that n×n = first n odd natural numbers. so in this way the 3 answers will be 16, 36 and 121 respectively
legendary sumit, is my ans right??
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Answers
Answered by
6
Let a be the first term and d be the common difference.
(1).
Here 1st term a = 1.
last term l = 9.
Number of terms n = 4.
We know that sum of n terms sn = n/2(a + 7)
= 4/2(1 + 7)
= 4/2(8)
= 4 * 4
= 16.
(2) Here a = 1, l = 11, n = 6.
We know that sum of n terms sn = n/2(a + l)
= 6/2(1 + 11)
= 6/2(12)
= 6(6)
= 36.
Therefore the sum = 36.
(3)
Method (1):
We know that sum of 1st n odd number = n^2
here total number of terms = 11.
Therefore the sum = (11)^2
= 121.
Method (2):
a = 1, n = 11, d = 2.
We know that nth term of an AP an = n/2 [2a + (n - 1) * d]
= 11/2[2(1) + (11 - 1) * 2]
= 11/2[2 + 10 * 2]
= 11/2[2 + 20]
= 11/2[22]
= 11 * 11
= 121.
Hope this helps!
(1).
Here 1st term a = 1.
last term l = 9.
Number of terms n = 4.
We know that sum of n terms sn = n/2(a + 7)
= 4/2(1 + 7)
= 4/2(8)
= 4 * 4
= 16.
(2) Here a = 1, l = 11, n = 6.
We know that sum of n terms sn = n/2(a + l)
= 6/2(1 + 11)
= 6/2(12)
= 6(6)
= 36.
Therefore the sum = 36.
(3)
Method (1):
We know that sum of 1st n odd number = n^2
here total number of terms = 11.
Therefore the sum = (11)^2
= 121.
Method (2):
a = 1, n = 11, d = 2.
We know that nth term of an AP an = n/2 [2a + (n - 1) * d]
= 11/2[2(1) + (11 - 1) * 2]
= 11/2[2 + 10 * 2]
= 11/2[2 + 20]
= 11/2[22]
= 11 * 11
= 121.
Hope this helps!
:-)
good man
Then y u reported my answer
because you're not wrote it in a perfect Manner
Ohkkk...Thanks
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When you dont like the answer, tell directly..No need to report...
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Answered by
2
we know that, for any natural number n,
n×n = sum of first n natural numbers.
this is a special property of rational numbers
Keeping this property in mind, we have
1) 1+3+5+7 = sum of first 4 natural numbers = 4×4 = 16
2) 1+3+5+7+9+11 = sum of first 6 natural numbers = 6×6 = 36
3) 1+ 3 +5+7+9+11 +13+15+17+19+21 = sum of first 11 natural numbers= 11 × 11 = 121
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