If the sum of n terms of an A.P. is nA + {n^2 | vedclass

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Question

(d) Given that $S_n=nA + {n^2}B$

Putting $n = 1,\;2,\;3,\;.............,$ we get,

${S_1} = A + B,\,{S_2} = 2A + 4B,\,\,{S_3} = 3A + 9B$

There

Vedclass · Accepted Answer

$2B$

Vedclass · Answer

(d) Given that $S_n=nA + {n^2}B$

Putting $n = 1,\;2,\;3,\;.............,$ we get,

${S_1} = A + B,\,{S_2} = 2A + 4B,\,\,{S_3} = 3A + 9B$

Therefore ${T_1} = {S_1} = A + B,\;{T_2} = {S_2} - {S_1} = A + 3B,$

${T_3} = {S_3} - {S_2} = A + 5B$,

Hence the sequence is $(A + B),(A + 3B),\;(A + 5B),...$

Here $a = A + B$ and common difference $d = 2B$.

If the sum of $n$ terms of an $A.P.$ is $nA + {n^2}B$, where $A,B$ are constants, then its common difference will be

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