find the sum of all 2 digit odd positive number?
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Answered by
601
Hi frnd here is ur ans,
Two digit odd positive numbers are 11,13,15,17...........99 are in A.P.
Here a = 11 and d = 2, tn= 99, n = ?
Sum of the n terms = (n/2)[2a+(n -1)d]
But tn = a + (n -1)d
⇒ 99 = 11+ (n-1)2
⇒ 99 -11 = (n-1)2
⇒ 88/2 = (n-1)
∴ n = 45.
subsitute n = 45 in sum of the n terms we obtain
⇒ s45 = (45/2)(2×11 + (45 -1)2)
⇒ s45 = (45/2)(110)
⇒ s45 = 45×55.
⇒ s45 = 2475.
∴sum of all two digit odd positive numbers = 2475.
...i hope this will help you pls mark as brainliest ☺
Two digit odd positive numbers are 11,13,15,17...........99 are in A.P.
Here a = 11 and d = 2, tn= 99, n = ?
Sum of the n terms = (n/2)[2a+(n -1)d]
But tn = a + (n -1)d
⇒ 99 = 11+ (n-1)2
⇒ 99 -11 = (n-1)2
⇒ 88/2 = (n-1)
∴ n = 45.
subsitute n = 45 in sum of the n terms we obtain
⇒ s45 = (45/2)(2×11 + (45 -1)2)
⇒ s45 = (45/2)(110)
⇒ s45 = 45×55.
⇒ s45 = 2475.
∴sum of all two digit odd positive numbers = 2475.
...i hope this will help you pls mark as brainliest ☺
doreamon1:
can you mark my ans as brainliest
Yep 2475 is the right answer...
hmm ss
Thanx
welcome
Answered by
85
Answer:2475
Step-by-step explanation:
11,13,15,17..........99
a=11
d=13-11
=2
l =99
n=?
l=a+(n-1)d
99=11+(n-1)2
99=11+2n-2
99=9+2n
99-9=2n
90=2n
n=90/2
n=45
Sn=n/2(a+l)
S45=45/2(11+99)
S45=45/2(110)
S45=45×55
S45=2475
Therefore, sum of all two digit odd positive number are 2475
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