1yes ans is b) 9998
The 99th term of the series 2+7+14+23+34+....... is
a)9999
b)9998
c)10000
d)none
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62For such questions, you should know the method of differences..
See, in an AP, the first difference is constant..
so the sum of n terms is a Quadratic.
If the 2nd difference is constant, then you can say that the sum of n terms will be a cubic
if the 3rd difference is constant then you can say taht the sum of n terms will be a forth degree polynomial...
and so on...
Now using this idea, we can say that
The sum of n terms here is a cubic so of the form
an3+bn2+cn+d
What remains is to find the constant a, b, c and d
moreover, if you substitute n=0 the sum is zero.. that means d=0
Now you have 3 variables, a, b, and c
Find the values by solving three simultaneous equations...
This may not be the best method here but a really useful one to know :)
1yupss me too[132][132][132][132][132][132][132][132][132][132][132][132][132][132][132]
1Ans (b).
let the terms be t1 , t2 , t3 , ....
so t2 - t1 = 5 which is same as (2x2)+1
t3 - t2 = 7 (2x3)+1
t4 - t3 = 9 (2x4)+1
---------------
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t99 - t98 = (2x99)+1 = 199
so adding the terms :
t99 - t1 = 9996
t1 =2
so t99 = 9998
so (b) ans.
1Sn=2+7+14+23+34+.................+tn
Sn= 2+7+14+23+34+.................+tn
subtracting we get
0=[2+5+7+9+11+.............] - tn
tn=2+[5+7+9+11+...............]
now tn=2+[(1+3+5+7+9+11+...............)-1-3]
sum of first n odd numbers=n2
so, tn=2+n2-4
tn=n2-2
iske baad mene n=100 kar diya
then tn=9998
1the ans is b
the second order difference is 2 so the nth tern is of the form an^2 +bn+c
now put n=1,2,3 and solve the 3 equations to det the values of a,b,c as 1,2,-1 respectively
so nth term=n^2+2n-1. now solve for n=99.
