@gs4890 said:is it correct ? will explain then.
yes.
@vbhvgupta said:yes.
dekho , try solving such questions by assuming values.
let the series be => 8,4,0,-4,-8,.....
now , ath term is 8 & bth term is 4
thus, (a+b)th term is 0.
there is another approach but a bit lengthy.
let the series be => 8,4,0,-4,-8,.....
now , ath term is 8 & bth term is 4
thus, (a+b)th term is 0.
there is another approach but a bit lengthy.
@sunnychopra89 said:In a 50 digit +ve no., all the digits are 4 except for the nth digit. This no. is divisible by 13 for some choice of that nth digit. How many possible values can n have?
10^6 mod 13 is 1
case1)
44+n44444 mod 13
5+n*10^5+ 10
5+4n+10
4n+15
4n+2
n=8
case2)44+4n4444 mod 13
5+16+3n+11 mod 13
3n+6 mod 13
n=11 not possible
case3)44+44n444 mod 13
5+16+12-n+2 mod 13
35-n mod 13
9-n mod 13
n=9
case4)
44+444n44
5-444+9n+5
8+9n
n=2
case5) 44+4444n4
5+4444*9-3n+4
5-18+4-3n
-3n-9
10n+4
n=10 not possible
case6)
44+44444n
5+9n
not possible
so n can be 6kth position ,6k+4,6k+3 among the first 48 positions from the left
24 cases
case7)4n+444444
4n +0
n not possible
case 8)n4 +444444
n4 +0
not possible
hence total 24 cases
@gs4890 said:dekho , try solving such questions by assuming values.let the series be => 8,4,0,-4,-8,.....now , ath term is 8 & bth term is 4thus, (a+b)th term is 0.there is another approach but a bit lengthy.
Couldn't get...it shd be 12...
@vbhvgupta said:Couldn't get...it shd be 12...
ath is the 1st term
bth is the 2nd
thus, (a+b)th i.e. 3rd term acc. to the series is 0
bth is the 2nd
thus, (a+b)th i.e. 3rd term acc. to the series is 0
@Subhashdec2 said:10^6 mod 13 is 1case1)44+n44444 mod 135+n*10^5+ 105+4n+104n+154n+2n=8case2)44+4n4444 mod 135+16+3n+11 mod 133n+6 mod 13n=11 not possiblecase3)44+44n444 mod 135+16+12-n+2 mod 1335-n mod 139-n mod 13n=9case4)44+444n445-444+9n+58+9nn=2case5) 44+4444n45+4444*9-3n+45-18+4-3n-3n-910n+4n=10 not possiblecase6)44+44444n5+9nnot possibleso n can be 6kth position ,6k+4,6k+3 among the first 48 positions from the left24 casescase7)4n+4444444n +0n not possiblecase 8)n4 +444444n4 +0 not possiblehence total 24 cases
Bhai kuch nei samjha....
What is the no. of +ve integer triplets(a,b,c) that satisfy abc=4(a+b+c) with a
@sunnychopra89 said:What is the no. of +ve integer triplets(a,b,c) that satisfy abc=4(a+b+c) with a
no solution??
+@vbhvgupta said:find the value of 1-2-3+2-3-4+..+ upto 100 terms.
1] 1-2-3 = -4
2] 2-3-4 = -5
we see a pattern
summation of this will be -[33/2(4 + {33-1}1] = -650
also we see that 100th term will be +34
therefore summation - -650+34 = -626
is this correct ?
@VaibhavJ said:1] 1-2-3 = -4 2] 2-3-4 = -5we see a patternsummation of this will be -[33/2(4 + {33-1}1] = -650also we see that 100th term will be +34therefore summation - -650+34 = -626is this correct ?
YES
@sunnychopra89 said:Bhai kuch nei samjha....
bhai parts mein pooch kaun kaun se step samajh ni aaye
saara samjhane betha to ho gya kaam
@sunnychopra89 said:OA is 24.....
44444...6 times is divisible by 13
lets group all the 50 digits into 8 groups of 6 digits and the remaining two.
if x is not one of the last two digits in that case,
case i) x44444 + 44 mod 13=0=> x=6
case ii) 4x4444 + 44 mod 13=0=> x=no value
case iii)44x444 + 44 mod 13=0=> x=9
case iv)444x44 + 44 mod 13=0=> x=2
case v)4444x4 + 44 mod 13=0=> x=no value
case vi)44444x + 44 mod 13=0> x= no value
if x is one of the last two digits no such value is possible.
so total no of values possible=8*3=24
@vbhvgupta 1-2-3 + 2-3-4 + 3-4-5 + .... 100 terms
1 + 2 + 3 + ..... 33 - ( 2+3+4+5....34 ) - (3+4+5+6+.... 35) + 36
=> 33*34/2 - ( 34*35/2 - 1 ) - ( 35*36/2 - 3 ) + 36
=> 561 - 595 + 1 - 630 + 3 + 36
=> -34 - 630 + 40
=> -624 answer...
@sunnychopra89 said:What is the no. of +ve integer triplets(a,b,c) that satisfy abc=4(a+b+c) with a
None.. ??
@vbhvgupta said:THe sum of first 4 terms of ap is 28 and sum of first 8 term of the same ap is 88. Find the sum of first 16 terms of the ap?
@Logrhythm can you solve this?
@vbhvgupta said:Q
1080 m ?
120+120*2*4/5+120*2*(4/5)^2+............
120+2*120*4/5(1/1-(4/5))
120+240*4=1080 m