find the sum of ....?

Question

Find the sum of all two digit odd positive numbers

Answer 1

2 digits....

i.e. 11 + 13 + 15 +..........+ 99


Last term is 99 and first term is 11....d = 2

nth term = a + (n-1)d = 99

11 + (n-1)2 = 99
(n-1)2 = 88
n-1 = 44
n = 45

Formula for Sum on n terms for an AP is

S = n/2 X {2a +(n-1)d}

S = 45/2 X {22 + 88}
S = 45/2 X 110 = 2475

Answer 2

There are 90 two digit numbers of which half is odd. 1,3,5,7,9 Which is 5 numbers for each decade or 45 numbers that are odd.
Taking the first and last odd number one can see that some is 110. Take the next two numbers and the sum is the same.
11 +99 = 110
13 + 97 = 110

110 x 44/2 + the middle odd number in the series. 55 = 2420 + 55 = 2475

I see there is more than one way to skin a cat.

Answer 3

This is an arithmetic progression. No. of terms = 45. Difference between two terms is 2. first term is 11. So use the summation formula of A.P and we get the sum as 2475.

Answer 4

woah...you really have to do that?
Is it that you don't know how to do it, or you want someone else to do it for you?
Just start with 11 and add every number up to 99 that is an odd number.
I would recommend using a calculator, if that is allowed.

Okay, I tried it on my calculator and got 2475...but I'm not rechecking it, so I'm not sure if that is right. IF you try and get the same, it is probably correct!

Answer 5

11+13+15...=50 +25
100+25
150+25
200+25; 250+25
300+25 350+25+400+25+450+25
=(50+100+150...)+25*9etc.
a=11, d=2
t1=11 + 2(i-1) 99=11+2(i-1); 88=2(i-1); 44=i-1; i=45
sn=n[2a+d(n-1)]/2; s45=45[2(11)+2(44)]/2
=2475

Answer 6

Where n is greater or equal than 6 and lower or equal to 50,
sum of (2n-1) = 2475

Answer 7

2,475 is the answer.

First, sum the ones' digits as (1+3+5+7+9)*9 = 225

Second, sum the ten's digits as 50*(1+2+...+9) = 50 (9) (8) = 2,250

225 + 2,250 = 2,475

Answer 8

2475

11+13+15+...+99=S
99+97+95+...+11=s

(110)(45)=2s

s=2475

Answer 9

2475