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ER. Gokul Krishnan Profile page        View Public Wall My Profile Page

Gokul Krishnan to kriti kankariya    6 years ago
Good Morning My sister...Can you please tell me what is sap? and for what it is used?
Reply Like? Yes (2) | No   
Gokul Krishnan to JANARTHANAN RAM    6 years ago
23.1 square + 48.6 square - 39.8square =x+1147.69
by unit digit method,
in this problem 1+6-4-9 = x which is equal to x = -6 then how can it equal to x =4
Reply Like? Yes (5) | No   
Gokul Krishnan to JANARTHANAN RAM    6 years ago
A can do a piece of work 4 hour, B and C can do it In 3 hour while A and C can together do it in 2 hour how long will B alone take to do it ?
Reply Like? Yes (2) | No (1)  
neha gupta   6 years ago
B alone will take 12 hours to do it.
Like? Yes (1) | No   
Gokul Krishnan to JANARTHANAN RAM    6 years ago
23.1 square + 48.6 square - 39.8square =x+1147.69
by unit digit method,
in this problem 1+6-4-9 = x which is equal to x = -6 then how can it equal to x =4
Reply Like? Yes (1) | No   
Gokul Krishnan to JANARTHANAN RAM    6 years ago
How the unit digit of (137^13)^47 is 3 ?
Brother,I am so puzzled,Please help me without fail...
I am getting the answer of 1....Please help me.I am waiting for your reply...
Reply Like? Yes (1) | No   
Saurabh Chaudhary   6 years ago
137^13 can be written as 7^13,cyclicity of 7 is 4,divide 13 by 4 you get remainder 1 therefore 7^1.Now solve for 7^47 again divide by 4 you get remainder 3,now 7^3=343,hence unit digit is 3.
Like? Yes (1) | No   
SunNy Yadav   6 years ago
Unit digit 3 hi hoga ...
As we know
{(137)^13}^47 = (137)^611

Now according to rule.. Kyunki Base ka unit 7 hai datsy ...
1) 611 - 1 =610
2) 610÷4 = remainder ayega 2
3) 2+1 = 3
Now Base ke unit ki power 3 ko bana do aur jo unit aye result me wahi hoga ..
As 7^3 = 7×7×7 = 343
Unit = 3
Like? Yes (1) | No   
SunNy Yadav   6 years ago
Rule =

Agar Base me unit 0, 1, 5, 6 ho to use solve karne par unit bhi 0, 1, 5 ya 6 hi hoga ...

But agar base ke unit me 2, 3,4,7,8,9 ho to..
Power me pehle 1 minus karo
Fir use 4 se divide karo
Jo remainder aye usme 1 plus karo
Then base ki unit me woh power rakh do..
Jo result me unit ayega wahi hoga ..
Like? Yes (1) | No   
Gokul Krishnan to JANARTHANAN RAM    6 years ago
The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest p.c.p.a?
Reply Like? Yes (1) | No   
Gokul Krishnan   6 years ago
The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest p.c.p.a?
Like? Yes (1) | No   
JANARTHANAN RAM   6 years ago
Ans: 15%

The difference b/w the compound interest and the simple interest over 2 years=P*(r/100)^2
P*(r/100)^2=405;
18000(r/100)^2=405;
r^2=405*10000/18000=225
r=15 %
Like? Yes (2) | No   
Gokul Krishnan to JANARTHANAN RAM    6 years ago
Brother when the series is squared,then we use the formulae
(n(n+1)(2n+1))/6, how you changed the formulae...help me, still I am very much confused....
Reply Like? Yes (1) | No   
JANARTHANAN RAM   6 years ago
Hi bro,
1^2 + 3^2 +5^2 +.....,n term =(n/3)*(2n+1)*(2n-1)------(1)
2^2 + 4^2 +6^2 +.....,n term =(2n/3)*(n+1)*(2n+1)------(2)
For even,
(1^2 - 2^2 +3^2 -4^2+......n term)= (1^2 + 3^2 +5^2 +...n/2 term) - (2^2 + 4^2 +6^2 +.....,n/2 term)
Using (1) & (2)
(1^2 - 2^2 +3^2 -4^2+......n term)= -n*(n+1)/2
For Odd,
(1^2 - 2^2 +3^2 -4^2+......n term)= (1^2 + 3^2 +5^2 +...(n+1)/2 term) - (2^2 + 4^2 +6^2 +.....,(n-1)/2 term)
Using (1) & (2)
(1^2 - 2^2 +3^2 -4^2+......n term)= n*(n+1)/2
Like? Yes (2) | No   
Gokul Krishnan   6 years ago
The difference between the compound interest and the simple interest accrued on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest p.c.p.a?
can you help me to solve in the easiest way?
Like? Yes (1) | No   
Gokul Krishnan to Pargat Pal Singh    6 years ago
Brother could you please find the answer of this question find the sum of first 40 terms of series 1^2-2^2+3^2-4^2+...
Reply Like? Yes (2) | No   
Gokul Krishnan to Rats    6 years ago
Brother could you please find the answer of this question find the sum of first 40 terms of series 1^2-2^2+3^2-4^2+...
Reply Like? Yes (2) | No   
Gokul Krishnan to Karthik Pandiyan    6 years ago
Brother could you please find the answer of this question find the sum of first 40 terms of series 1^2-2^2+3^2-4^2+...
Reply Like? Yes (1) | No   
Gokul Krishnan to C.T.Narayanan    6 years ago
Brother could you please find the answer of this question find the sum of first 40 terms of series 1^2-2^2+3^2-4^2+...
Reply Like? Yes (2) | No   
Gokul Krishnan to MOJUMDER    6 years ago
Brother could you please find the answer of this question find the sum of first 40 terms of series 1^2-2^2+3^2-4^2+...
Reply Like? Yes (1) | No (1)  
MOJUMDER   6 years ago
(40*41*81)/6 - 2*2^2[(20*21*41)/6] = -820. Think.
Like? Yes (2) | No   
Gokul Krishnan to RAKESH    6 years ago
Brother could you please find the answer of this question find the sum of first 40 terms of series 1^2-2^2+3^2-4^2+...
Reply Like? Yes (1) | No   
RAKESH   6 years ago
(1^2-2^2)+(3^2-4^2)+(5^2-6^2)... +(39^2-40^2)
a^2-b^2=(a+b)(a-b)
= - [1+2+3+4+5+6+---------+ 39+40]
= -20*41
= -820
Like? Yes (2) | No (1)  
Gokul Krishnan to susmit    6 years ago
Brother could you please find the answer of this question find the sum of first 40 terms of series 1^2-2^2+3^2-4^2+...
Reply Like? Yes (1) | No (1)  
Gokul Krishnan to ~Sandeep~    6 years ago
Brother could you please find the answer of this question find the sum of first 40 terms of series 1^2-2^2+3^2-4^2+...
Reply Like? Yes (3) | No (1)  
Gokul Krishnan to Dr.Dipin Singh    6 years ago
Brother could you please find the answer of this question find the sum of first 40 terms of series 1^2-2^2+3^2-4^2+...
Reply Like? Yes (4) | No (1)  
Gokul Krishnan to JANARTHANAN RAM    6 years ago
Brother could you please find the answer of this question find the sum of first 40 terms of series 1^2-2^2+3^2-4^2+...
Reply Like? Yes (1) | No   
JANARTHANAN RAM   6 years ago
Hi Bro,
Sum of 1^2-2^2+3^2-4^2+...+n^2= - (n/2)*(n+1) (If n is even)
So Sum= - (40/2)(40+1)= -820
Like? Yes (6) | No   
Gokul Krishnan   6 years ago
Thanks My Brother...What about when the N is odd condition (ie) what is the formulae?
Like? Yes (1) | No   
JANARTHANAN RAM   6 years ago
Sum of 1^2-2^2+3^2-4^2+...+n^2= - (n/2)*(n+1) (If n is even)
Sum of 1^2-2^2+3^2-4^2+...+n^2= (n/2)*(n+1) (If n is Odd)
Like? Yes (1) | No (1)  
veeraiah    8 years ago
hai..............
Reply Like? Yes | No   
  • ER. Gokul
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Gokul Krishnan Tamil Nadu
Gokul krishnan (M)
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M4 Score
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