1))fIND THE SUM upto n terms-2 +12+36+80+150+252.....
2))if a,b,c, r in HP then bc,ca,ab r in?
30I'll give hints only ---
1) a) By the time we reach the nth bracket, 1+2+3+...+(n-1) terms would have been used.
First term of nth bracket= n(n-1)2+1
b) Apply the previous logic here too.
2) arrange it like this
Now count how many times each digit appears in hundredth, tens and units places. Its not too difficult and the result will be symmetric for all digits 1-9. You dont need to rack your brains for 0 because it doesn't add to the sum.
I think the answer is 12600
3)
1s=1+2*2+3*22......100*299
2s= 2+2*22+3*23....100*2100
s-2s=1+2+22+23.......-100*2100
-s=1+2100-12-1-100*2100
s=99*2100
262take S, then take 2S and then subtract them , then just normal Gp
7Find S=1+2x2+3x22+4x23.............100x299.
S means to find the sum.
x means to multiply.
113)
S=1+12-14-18+116......
S=1+12-14(1+12-14...)
S=1+12-S4
5S4=32
S=65...Answer..;)
71)Find the sum of the terms in th nth bracket:
a))(1), (2,3), (4,5,6), (7,8,9,10) ........
b)) (12) (22,32) (42,52,62).......
2)Find the sum of the sum of the digits of all 3 digit nos.
3)Find S=1+1/2-1/4-1/8+1/16+1/32-1/64-1/128..................
11@jit-
The series goes like this 2,12,36,80...till n..which follows the form of n3+n2
13+12=2.
23+22=12.
33+32=36
.
.
.
.
n3+n2
Hence the sum of the series will be equal to the Σ n3 +Σ n2..
Hope u get it now..:):):)
1if a,b,c are in hp
then 1 ,1 ,1 are in A.P
a b c
by multiplying all three wth abc we get bc,ca and ab → which implies bc ca and ab r in A.P;-)
11The series is in the form n3+n2
[n(n+1)2]2 +n(n+1)(2n+1)6
n(n+1)2{n(n+1)2 + 2n+13} Answer..:)
112)
1a,1b,1care in AP
1b-1a=1c-1b
a-bab=b-cbc
ca-cb=ab-ac
therefore bc,ca and ab are in AP..:)
