Let S_{n} be the sum of the first n terms of an arithmetic progression. If S_{3n} = 3S_{2n}, then the value of \frac{S_{4n}}{S_{2n}} is

Question

Let S_n be the sum of the first n terms of an arithmeticn progression. If S_3n = 3S_2n, then the value ofn fracS_4nS_2n is

DivineJEE · Accepted Answer

nnS_n=(n / 2)(2 n+(n-1) d)nnnnthe formula we can mopmers S_ lend S_ ne nnnn nnS_n n=(3 n / 2)(2 n+(3 n-1) d) nnS_n n=(2 n / 2)(2 n+(2 n-1) d)=n(2 n+(2 n-1) d)nn nnnn3. "Set Up the Given Equation": Accondingts thaprablim, we have:nnnn5_m=15_mnnnnnn(3 n / 2)(2 n+(3 n-1) d)=3(n(2 n+(2 n-1) d))nnnn4. "Simplity tha Equation": Owldingbethsidasby n(maxumingn / ©:nnnn(3 / 2)(2 n+(3 n-1) d)=3(2 n+(2 n-1) d)nnn nNow, multiply both aldou by 2 todelirrinsto the fraction:nnnn3(2 a+(m n-1) d^2)-6(2 a+(2 n-1) d)nnnn5. "Bepard Both Sidea": Equarding both aldes glve:nnnn6 x+9 n d-3 d=15 n+6 n d-6 dnnnn5. "RearrangingTurra": Rearrangingthesquation:nnnn6 a+9 n d-1 d-13 n-6 n d+6 d=0nnn nElmpilinging further:nnnn-8 n+3 n d+3 d=0nnn nDividingthroughtry 3:nnnn-5 a+ md+d=0nnn nThus, we hind:nnnn2 a=n d+d > 2 n=d(n+1)nnnnnnS_ m=frac4 n2( m+(4 n-1) d)= mn( m n+(4 n-1) d)nnnnB. "Qubeituting la^21, Subeituta 2 a= d( x+1) irta S_ axnnnnS_n=ln (d(n+1)+(4 n-1) d)=2 n(d(n+1+4 n-1))=ln (d(5 n))=10 n d^2nnnn7. "Ninding S. S. Now, we calculate Shicnnnn5 m=n(2 n+(2 n-1) d^prime)=n(d(n+1)+(2 n-1) d)=n(d(n+1+2 n-1))=n(nnn nThandfore, weonhed theratio:nnnn(S_m / S_m)=(10 n d^2 / 3 n E^1)=(10 / 3)nnnnnnfracS_n xS_n x=(10 / 3)nnn

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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