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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?

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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

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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 ?

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neha gupta 6 years ago

B alone will take 12 hours to do it.

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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 =4ï»¿

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...

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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.

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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) 610ÃƒÂ·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 = 7Ã—7Ã—7 = 343
Unit = 3

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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 ..

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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?

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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?

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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 %

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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....

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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

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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?

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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+...

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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+...

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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+...

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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+...

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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+...

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MOJUMDER 6 years ago

(40*41*81)/6 - 2*2^2[(20*21*41)/6] = -820. Think.

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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+...

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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

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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+...

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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+...

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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+...

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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

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Gokul Krishnan 6 years ago

Thanks My Brother...What about when the N is odd condition (ie) what is the formulae?

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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)

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veeraiah 8 years ago

hai..............

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ER. Gokul
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Gokul krishnan (M)

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