IIT JEE , AIEEE Forum - Mathematics , Algebra - this question is about arithmetic series. i have got the solution. please explain me it.

krithika

If S_m : S_n is equal to 441:256. what is the ratio of t_m: t_nequal to?

your options are:

a) 21:16 b) 41:31 c) 31:41 d) 16:21

the formulae to be used in this sum are:

S_n= n/2 (2a + (n-1)d) and t_n= a + (n-1)d

please reply. points will be given for sure.give the correct answer with proper explained solution.

the solution is: i have found the solution. please explian me the soltuion alone.

Sm=441=21^2=Sum of first 21 odd numbers(Sum of first n odd numbers is n^2).Also Sn= 256=Sum of first 16 odd numbers.So Tm=(2*21)-1=41 and Tn=(2*16)-1=31.So the answer is 41:31

just clarify me. how t_nand t_m are being found? please. what formulae is used here?

nisha_07

I READ SOMEWHERE DAT

S_m/S_n=(t₁/t₂)²

nisha_07

IS THE ANSWER a.21:16

krithika

satyabharadwaja

I think you are little bit confused about it
It is a standard result that sum of the the n odd numbers is n^2
then
Sm=441k=n/2[2a+(n-1)d]
Since Sm is the sum of the 21 odd numbers n=21
Here k is the constant of ratio
a=1 d=2
441k=21/2[2(1)+(21-1)2]
simplifying it we will get k=1
therefore calculating Tm and Tn is not a biggger task
as we know that a=1 d=2 and in
Tm (n=21)
Tn (n=16)
then the ratio will be of 41:31 itself

krishna.gopal

But how do you know that the series in this question is 1,3,5.7.......
For a general AP where a and d are not given no one can find the answer of this question. If a and d is given then find relation between m and n condition given for Sm/Sn and use it to find tm/tn.

satyabharadwaja

the standard result that n^2 is the sum of the first n odd numbers
justb we are considering it as one of the case

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