IIT JEE , AIEEE Forum - Mathematics , Algebra

Avinash_Bhat

An arithmetic progression consisting up of ' 6t ' positive real terms is given.

The sum of the first ' 2t ' terms is given by : S1.

The sum of the first ' 4t ' terms is given by : S2.

The sum of the last ' 2t ' terms is given by : S3.

If (S1.S2.S3) - (S1².S3) = 4

then (S2 - S1) is not less than :

a) 2^3/2

b) 2^2/3

c) 2^1/3

d) 2^5/2You need to explain your answer :

Thathwamasi

r_age

is it 2 raised to 1/3

Avinash_Bhat

SORRY !!! Wrong answer

r_age

i dont know,do you have solution?

Avinash_Bhat

Now I edit the quetion asking you to explain your answer. If you are perfect, I will rate you with the best.

sunayana_kushwaha

hey is it S1 raise to power s2 raise to power s3 or s1.s2.s3??

swat3cat

i think its 2 raised to 3/2

Avinash_Bhat

Hey Sunayana, its (S1.S2.S3) - (S1².S3) = 4
I have now edited it....

And..... Hey SWAT, you need to explain your method of approach, then only I can say whether it is right or wrong.

vish0001

hey answer is comin as (b) = 2 to the power 2/3

vish0001

here's the solution

As per the qn, there are 6t terms in an A.P. S1 : sum of first 2t terms

S2 : sum of first 4t terms

S3 : sum of last 2t terms

given 3n terms of an A.P.,

Sum of 1st n terms, Sum of next n terms, Sum of last n terms are in A.P.

Then here S1, (S2 - S1), S3 are in A.P. ---------- (1)

Also given, (S1*S2*S3 - S1^2*S3) = 4;

ie, S1*(S2 - S1)*S3 = 4

A.M. >= G.M.

{ S1 + (S2 - S1) + S3 } / 3 >= cuberoot(S1*(S2 - S1)*S3)

>= cuberoot(4)

From eqn(1) 3(S2 - S1) / 3 >= cuberoot(2^2)

(S2 - S1) >= 2^2/3

SO : (S2 - S1) not less than 2^2/3 .........

vish0001

hey avinash is it right ??

Avinash_Bhat

You are right............