Dear friends, please solve the following.
1. let the four terms be 5+d, 5+2d,5+3d,5+4d. so 230 is 5+5d implying that d is 45 so the four terms are 50,95,140,185
2. 2*(6x+2) = 4x+4+8x
we get 12x =12x which is an identity which implies that x can be any number and the sequence will always be in A.P.
3. sum of squares of first twelve numbers = n*(n+1)*(2n+1)/6 = 12*13*25/6 = 650
4. let nth term be represented by t(n)
t(n) - t(n-1) = 2*n -1
t(2) - t(1) = 3
t(3) - t(2) = 5
t(4) - t(3) =7
.......
t(n) - t(n-1)= 2*n - 1
t(n) - t(1) = sigma(2*n -1) from n =2 to n =n
the above expression is equal to ((sigma(2*n-1)from n=1 to n=n ) - 1)
i.e. n square -1 (easier way to do the 4th one is to just sunstitute n=3 in all the options and see whther you get 8 )
a). 5----230
n=6,d=-1,a=5
t4=230
a+(n-1).d=230
5+(4-1).d=230
5+3d=230
3d=125
d=41.67
=42
we know,
a,a+d,a+2d,a+3d
5,5+42,5+82,5+126
5,47,87,131
b). we know,
a+c/2=b
4x+4+8x/2=6x+2
12x=12x
c). Sn=n/2(2a+(n-1)d)
S12=12/2(2*1+(12-1)*3)
=6(2+11*3)
=6(2+33)
=210 ans.......
d). 2n^2-1
CAclubindia
