find a solution

[catquery]

can u explain for the second question how did you get option c because i am getting option b i.e. 2.5 or 27.5

Originally Posted by Apprentice

Easy i guess...

1-ans is 506 the max sum is the sum of nos till 2, or IOW sum of first 22 even nos...
=n(n+1),=22*23=506

2-tried thru options, is the answer c-1.5..????

[popeye]

Originally Posted by catquery

can u explain for the second question how did you get option c because i am getting option b i.e. 2.5 or 27.5

The answer is option b ie 2.5/27.5. Here is the method

Let the first term of both AP & GP be a & common difference of AP be d & common ratio of GP be r. Hence we have
Series 1st term 2nd term 3rd term
AP a a+d a+2d
GP a ar ar^2

Now it is given a=2 & also 3rd terms are equal.

So we get

2+2d=2r^2
1+d=r^2...........(1)

Also from given condition of 2nd terms
a+d=ar + 0.25
2+d=2r+0.25
Substituting eqn 1 we get
r^2 + 1= 2r + 0.25

Solving we get
r=1/2 or 3/2
So we get d=-1/4 or 5/4

So substituting the 2 values of d in AP you can calculate the sum of first 5 nos which is either 2.5for d=-1/4 & 27.5 for d=5/4.

[thyisthyname]

me got a good question

units digit of 7^11^12^13

[catquery]

till finding the value of r it is ok but after that i think there is some correction

Originally Posted by popeye

The answer is option b ie 2.5/27.5. Here is the method

Let the first term of both AP & GP be a & common difference of AP be d & common ratio of GP be r. Hence we have
Series 1st term 2nd term 3rd term
AP a a+d a+2d
GP a ar ar^2

Now it is given a=2 & also 3rd terms are equal.

So we get

2+2d=2r^2
1+d=r^2...........(1) [ 1+d=r^2 or 1+d=1/4 or d= -3/4 ] [r=1/2 or r^2=1/4 ]

Also from given condition of 2nd terms
a+d=ar + 0.25
2+d=2r+0.25
Substituting eqn 1 we get
r^2 + 1= 2r + 0.25

Solving we get
r=1/2 or 3/2
So we get d=-1/4 or 5/4 [ d = - 3/4 or 5/4 ]

So substituting the 2 values of d in AP you can calculate the sum of first 5 nos which is either 2.5for d=-1/4 & 27.5 for d=5/4. [ d=5/4 then series= 2, 3.25, 4.5, 5.75, 7 and its sum will be 22.5 not 27.5 ]

[popeye]

Originally Posted by thyisthyname

me got a good question

units digit of 7^11^12^13

Lord Krishna honing his quant skills before joining NM

Here is my metod

Take 12^13. It will give last digit as 8 i.e even no.
Now 11^x always gives last digit 1 i.e an odd no.

so 11^12^13 will give an odd no. ending in 11

so 7^odd no. will give last digit as either 7 or 3 according to power cycle of 7

now 7^11^1 will give last digit 3
also 7^11^2 will give last digit7

continuing like this 7^11^ even digit will give last digit 7.

Hence according to this logic last digit sjould be 7.
Is it the answer, my lord ???

[popeye]

Originally Posted by catquery

till finding the value of r it is ok but after that i think there is some correction

Thanks for pointing out the error. In aste I did not calculate the sum.
So than none of the option satisfies the answer or else option b should be 2.5 or 22.5

[Apprentice]

Originally Posted by popeye

Thanks for pointing out the error. In aste I did not calculate the sum.
So than none of the option satisfies the answer or else option b should be 2.5 or 22.5

Rite..I was banging my head over the secund option yaar ..cos i tried for 27.5 ..dint get the ansewer...moved on n saw that 1.5 gave me the closest answer, cos the difference in 2 nd terms of AP n GP was closest to 0.25....

But ya u r rite...


thanx

[thyisthyname]

Originally Posted by popeye

Lord Krishna honing his quant skills before joining NM

Here is my metod

Take 12^13. It will give last digit as 8 i.e even no.
Now 11^x always gives last digit 1 i.e an odd no.

so 11^12^13 will give an odd no. ending in 11

so 7^odd no. will give last digit as either 7 or 3 according to power cycle of 7

now 7^11^1 will give last digit 3
also 7^11^2 will give last digit7

continuing like this 7^11^ even digit will give last digit 7.

Hence according to this logic last digit sjould be 7.
Is it the answer, my lord ???

the answer is correct ...... the lord has spoken

[catquery]

Find the value of 1-2-3 + 2-3-4 +....+ upto 100 terms.

(a) -694 (b) -626 (c) -624 (d) -549

What will be the sum to n terms of the series 8 + 88 + 888 +....?

(a) 8(10^n - 9n)/81 (b) 8{10^(n+1) - 10 - 9n}/81 (c) 8{10^(n-1) - 10} (d) Not

Originally Posted by popeye

Thanks for pointing out the error. In aste I did not calculate the sum.
So than none of the option satisfies the answer or else option b should be 2.5 or 22.5

[Naresh Reddy]

Originally Posted by catquery

Find the value of 1-2-3 + 2-3-4 +....+ upto 100 terms.

(a) -694 (b) -626 (c) -624 (d) -549

What will be the sum to n terms of the series 8 + 88 + 888 +....?

(a) 8(10^n - 9n)/81 (b) 8{10^(n+1) - 10 - 9n}/81 (c) 8{10^(n-1) - 10} (d) Not

(1)
(1-2-3)+(2-3-4).....uptp 100 terms
= (-4)+(-5) upto 33 terms+ 34
=(-4-5-6 upto 33terms)+34= [-33*(4+36)/2 ]+34=-660+34=-626 ANS (b)

(2)
8+88+888 .... =8*(1+11+111..)=8/9*(9+99+999....)
=8/9*[ 10-1+100-1+1000-1...nterms]
=8/9*[10+100+1000..n terms -n*1]
=8/9*[{10*(10^n - 1)/(10-1)} -n]
=8/9*[{10^n+1 -10/9} -n]
=8{10^n+1-10 -9n}/81 ANS (b)