In an A.P. if `S_1=T_1+T_2+T_3+.....+T_n(nod d)dotS_2=T_2+T_4+T_6+.........+T_(n-1)` , then find the value of `S_1//S_2` in terms of `ndot`

In an A.P.if S_(1)=T_(1)+T_(2)+T3+...+T_(n) (n odd S_(2)=T_(1)+T_(3)+T_(5)+...+T_(n) then (S_(1))/(S_(2))=

In an A.P. if a=2, t_(n)=34 ,S_(n) =90 , then n=

In an A.P., if T_(1) +T_(5)+ T_(10) +T_(15)+ T_(20) + T_(24) = 225, find the sum of its 24 terms.

Let {t_(n)} is an A.P If t_(1) = 20 , t_(p) = q , t_(q) = p , find the value of m such that sum of the first m terms of the A.P is zero .

In a harmonic progression t_(1), t_(2), t_(3),……………., it is given that t_(5)=20 and t_(6)=50 . If S_(n) denotes the sum of first n terms of this, then the value of n for which S_(n) is maximum is

Suppose a series of n terms given by S_(n)=t_(1)+t_(2)+t_(3)+ . . . . +t_(n) then S_(n-1)=t_(1)+t_(2)+t_(3)+ . . . . +t_(n-1),nge1 subtracting we get S_(n)-S_(n-1)=t_(n),nge2 surther if we put n=1 is the first sum then S_(1)=t_(1) thus w can write t_(n)=S_(n)-S_(n-1),nge2 and t_(1)=S_(1) Q. The sum of n terms of a series is a.2^(n)-b . where a and b are constant then the series is

Suppose a series of n terms given by S_(n)=t_(1)+t_(2)+t_(3)+ . . .. +t_(n) then S_(n-1)=t_(1)+t_(2)+t_(3)+ . . . +t_(n-1),nge1 subtracting we get S_(n)-S_(n-1)=t_(n),nge2 surther if we put n=1 is the first sum then S_(1)=t_(1) thus w can write t_(n)=S_(n)-S_(n-1),nge2 and t_(1)=S_(1) Q. if the sum of n terms of a series a.2^(n)-b then the sum sum_(r=1)^(infty)(1)/(t_(r)) is

The Fibonacci sequence is defence by t_(1)=t_(2)=1,t_(n)=t_(n-1)+t_(n-2)(n>2). If t_(n+1)=kt_(n) then find the values of k for n=1,2,3 and 4.