in a sequence the sum of first n terms is given by Sn = nP+(1/2)n(n-1)Q, where P and Q are constants. Show that the sequence is AP. Find the first term, common difference and the 100th term.
According to the answer given
first term is P
common difference is Q
100th term is P+99Q
I am finding Tn= Sn+1-Sn = P + nQ
but if we put T(1) then first term is coming P+Q
T(100) = P+100 Q
Is the answer wrong or I did the sum wrongly ??

Question

in a sequence the sum of first n terms is given by Sn = nP+(1/2)n(n-1)Q, where P and Q are constants. Show that the sequence is AP. Find the first term, common difference and the 100th term.
According to the answer given 
first term is P
common difference is Q 
100th term is P+99Q 
I am finding Tn= Sn+1-Sn = P + nQ 
but if we put T(1) then first term is coming P+Q
T(100) = P+100 Q 
Is the answer wrong or I did the sum wrongly ??

cwrw238 · Answer

plug in n = 1 and calculate the result 
this will give you the first term

anonymous · Answer

so then the first term would be P+Q !! but the answer given is P (the first term)

hartnn · Answer

after a lot of thought, i realized that Tn = Sn+1 - Sn is incorrect! 
it'd be Tn = Sn - S (n-1)

hartnn · Answer

Sum of n terms - sum of n-1 terms gives you n'th term., logical right ??
now you'll get your required result :)

anonymous · Answer

$$Sn-Sn-1 = nP-\frac{ 1 }{2 }n^2Q-\frac{ 1 }{2 }nQ-[(n-1)p+\frac{ 1 }{2 }(n-1)(n-2)Q]$$

hartnn · Answer

+1/2 n^2 Q

hartnn · Answer

you got that error ? it should be +1/2 n^2Q instead of -1/2n^2Q you wrote...

anonymous · Answer

thanks @hartnn for pointing out the mistake  !! i got the answer !!

hartnn · Answer

welcome ^_^