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$99^{\text {th }}$ term of the series $2+7+14+23+34+\ldots \ldots$ is
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Verified Answer
The correct answer is:
$9998$
Let $S=2+7+14+23+34+\ldots . .+T_n \ldots(i)$
and $S=2+7+14+\ldots \ldots \ldots \ldots \ldots .+T_{n-1}+T_n \ldots(ii)$
From (i) and (ii), we get
$0=2+\left[5+7+9+11 \ldots \ldots . .+T_n-T_{n-1}\right]-T_n$
$T_n=2+\left[\frac{n-1}{2}\{2 \times 5+(n-2) 2\}\right]$
$\Rightarrow T_n=2+(n-1)(n+3)$
Now put $n=99$
$\Rightarrow T_{99}=2+98 \times 102=9998$
and $S=2+7+14+\ldots \ldots \ldots \ldots \ldots .+T_{n-1}+T_n \ldots(ii)$
From (i) and (ii), we get
$0=2+\left[5+7+9+11 \ldots \ldots . .+T_n-T_{n-1}\right]-T_n$
$T_n=2+\left[\frac{n-1}{2}\{2 \times 5+(n-2) 2\}\right]$
$\Rightarrow T_n=2+(n-1)(n+3)$
Now put $n=99$
$\Rightarrow T_{99}=2+98 \times 102=9998$
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