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If the sum of $n$ terms of an A.P. is $n A+n^2 B$, where $A, B$ are constants, then its common difference will be
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The correct answer is:
$2 B$
Given that $S_n=n A+n^2 B$
Putting $n=1,2,3, \ldots \ldots . . . . .$. , we get
$S_1=A+B, S_2=2 A+4 B, S_3=3 A+9 B$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
Therefore $T_1=S_1=A+B, T_2=S_2-S_1=A+3 B$,
$T_3=S_3-S_2=A+5 B$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
Hence the sequence is $(A+B),(A+3 B),(A+5 B)$
Here $a=A+B$ and common difference $d=2 B$ $\ldots\ldots. . . . .$
Putting $n=1,2,3, \ldots \ldots . . . . .$. , we get
$S_1=A+B, S_2=2 A+4 B, S_3=3 A+9 B$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
Therefore $T_1=S_1=A+B, T_2=S_2-S_1=A+3 B$,
$T_3=S_3-S_2=A+5 B$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\ldots\ldots\ldots\ldots\ldots\ldots. . . . .$.
Hence the sequence is $(A+B),(A+3 B),(A+5 B)$
Here $a=A+B$ and common difference $d=2 B$ $\ldots\ldots. . . . .$
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