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Numerical Ability
Geometry
87th number in the series 2, 10, 26, 50…….
Read Solution (Total 16)
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- Observing the given we can write the following :
2nd term - 1st term = 1*8;
3rd term - 2nd term = 2*8;
4th term - 3rd term = 3*8;
5th term - 4th term = 4*8;
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.
.
(n-1)th term - (n-2)th term =(n-2)*8
(n)th term - (n-1)th term =(n-1)*8;
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Summing up all the above terms we get ,
(n)th term - 1st term = {1+2+......+(n-1)}*8
=>(n)th term = 8*{n*(n-1)}/2 + 1st term
=>(n)th term = 8*{n*(n-1)}/2 +2;
So, Clearly 87th term of the sequence will be :
8*{n*(n-1)}/2 +2 = 8*{87*86}/2 +2 = 4*86*87 +2 = 29930
Hence, 87th term will be : 29930 - 12 years agoHelpfull: Yes(121) No(10)
- answer is =29930
series is 2,10,26,50
like 1st term is 2 = 1^2+1
2nd term is 10= 3^2+1
3rd term is 26= 5^2+1
similarly 87 term is = 173^2+1=29930......
as we seen above that in 1st ,2nd,& 3rd term it is squire of corresponding odd numbers+1
since 1st odd no is 1(1*2-1),,,,,2nd odd no is 3 (2*2-1),,,,3rd odd no is 5(2*3-1).....similarly 87 odd no is (2*87-1)= 173...... - 12 years agoHelpfull: Yes(76) No(4)
- Sorry for the wrong ans 29930 is correct.
when diff are in AP
Nth term = an^2 + bn +c
put a + b + c = 2
4a + 2b + c = 10
9a + 3b + c +26
a = 4
b = -4
c = 2
Lil trick i learnt in 12th
now put n = 87
ans = 29930 - 12 years agoHelpfull: Yes(16) No(4)
- 1st =2
2nd =2+1*8
3rd=2+1*8+2*8
4th=2+1*8+2*8+3*8
.....
....
nth=2+1*8+2*8+....+(n-1)*8, then
87th=2+1*8+2*8....(87-1)*8
87th=2+(87/2)*{2*1+(87-1)*8}
=2+29928
=29930 - 12 years agoHelpfull: Yes(12) No(0)
- ans is 690..
the gn series is of the form a,a+d,a+2d,a+3d....,a+nd...
here n=87
a+(n-1)d
2+(86*8)
-->690
- 12 years agoHelpfull: Yes(7) No(35)
- 1st =2
2nd =2+1*8
3rd=2+1*8+2*8
4th=2+1*8+2*8+3*8
.....
....
nth=2+1*8+2*8+....+(n-1)*8, then
87th=2+1*8+2*8....(87-1)*8
87th=2+(86/2)*{2*8+(86-1)*8}
=2+29928
=29930 - 12 years agoHelpfull: Yes(4) No(0)
- //sorry for previous wrong calculation
series is 2,10,26,50,...
to find t(87)
now 2, 10, 26, 50
common difference ---> 8 16 24
again c.d ----> 8 8
so T(n)=a(n-1)*(n-2)+b(n-1)+c
now put n=1,2,3
T(1)=c=2 ----(i)
T(2)=b+c=10 ----(ii)
T(3)=2a+2b+c=26--(iii)
from (i),(ii),(iii)
a=4,b=8,c=2
now u can calculate any term of the progression from above expression
T(n)=a(n-1)*(n-2)+b(n-1)+c
so T(87)=4*86*85+8*86+2=29930 Ans.
one can verify it by
T(5) = 4*4*3 + 8*4 +2
=82 - 12 years agoHelpfull: Yes(3) No(1)
- 2, 10, 26, 50
10-2=8
26-10=16
50-26=24
next no should be a 82
82-50=32 - 12 years agoHelpfull: Yes(2) No(20)
- series is 2,10,26,50,...
to find t(87)
now 2, 10, 26, 50
common difference ---> 8 16 24
again c.d ----> 8 8
so T(n)=a(n-1)*(n-2)+b(n-1)+c
now put n=1,2,3
T(1)=c=2 ----(i)
T(2)=b+c=10 ----(ii)
T(3)=2a+2b+c=26--(iii)
from (i),(ii),(iii)
a=4,b=8,c=2
now u can calculate any term of the progression from above expression
T(n)=a(n-1)*(n-2)+b(n-1)+c
so T(87)=4*86*85+8*86+2=688 Ans.
one can verify it by
T(5) = 4*4*3 + 8*4 +2
=82
- 12 years agoHelpfull: Yes(2) No(1)
- 2 = 12 + 1
10 = 32 +1
26 = 52 + 1
50 = 72 + 1
Now,
1,3, 5, 7 ...... forms a AP series with common difference 2.
So, we will find T87 firs through Ap.
Tn = a +(n-1)*d
Tn = 1 + (87 -1) *2 = 1 + 86 *2 = 173.
Now,
87th term = 1732 + 1 = 29929 + 1 = 29930. - 8 years agoHelpfull: Yes(2) No(0)
- a=2,d=8,n=87
a+(n-1)d
2+(86*8)
690 ans
- 12 years agoHelpfull: Yes(1) No(11)
- The difference = 8
8,16,24
the 87th difference is
8+(87-1)8
= 696
So 2-10-26
then 2+696 = 698 should be the ans , not sure though - 12 years agoHelpfull: Yes(1) No(10)
- the series is 2 10 26 50... from this series we got a series of multiple of 8
8 16 24 32
now we have 2 find 87th number in the 1st series means 86 in the 2nd series.
from the formula-> Tn=a+(n-1)*d
we got Tn=8+(85*8)=688
Now just sum the all terms from the formula Sn=(n/2)*[2a+(n-1)*d]
we got Sn=29928 and this is the answer - 12 years agoHelpfull: Yes(1) No(3)
- LALA JI Best solution... :)
- 9 years agoHelpfull: Yes(0) No(1)
- Its in the for ar^n
- 8 years agoHelpfull: Yes(0) No(0)
- The series is made up of addition of the previous term with multiples of 8.
First term: 2 + 8(0) = 2
Second term: First term + 8(1) = 2+8 =10
Third term = Second term + 8(2) = 10 + 16 = 26
Fourth term : Third term + 8(3) = 26 + 24 = 50
Can you see the pattern yet? We can generalize it to :
g(n) = g(n-1) + (n-1)(8)
If you try putting the values, such as n=3 (third term), you'll get
g(3) = g(2) + (2)(8). We know that g(2) (second term) is equal to 10.
Thus g(3) = 10 + 16 = 26.
The question is for 87th term.
g(87) = g(86) + 8(86)
But do we know g(86) ? No, we don't. But we can find it out.
Lets get back to the general equation.
g(n) = g(n-1) + (n-1)(8)
Substituite g(n-1) = g(n-2) + (n-2)(8)
Thus, g(n) = g(n-2) + (n-1)8 + (n-2)(8)
Similarly, substiute, g(n-2) = g(n-3) + (n-3)8
.
.
.
g(n) = g(1) + 8(n-1) + 8(n-2) + 8(n-3) + 8(n-4)..... 8(n-(n-1))
We didn't know the 86th term. But we do know the 1st term i.e. g(1) = 2
g(n) = 2 + 8[ n-1 + n-2 + n -3+..... n-n+1]
g(87) = 2 + 8 [86 + 85 + 84 + 83+....1]
1+2+3+4+....86 is a series with common difference = 1 and first term = 1. The sum of this type of series is n(n+1)/2 i.e (87)(86)/2.
Therefore,
g(87) = 2 + [8(86)(87)/2]
g(87) = 2 + [59856/2]
g(87) = 2 + 29928
g(87) = 29930 - 6 years agoHelpfull: Yes(0) No(0)
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