Quantitative aptitude

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