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(B) The sum of an arithmetic progression is given by $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.Given $d=1$,$\sum_{i=1}^{n} a_i = \frac{n}{2}[2a_1 + (n-1)] =

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$96$

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(B) The sum of an arithmetic progression is given by $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.Given $d=1$,$\sum_{i=1}^{n} a_i = \frac{n}{2}[2a_1 + (n-1)] = 192$.$2a_1 + n - 1 = \frac{384}{n} \quad \dots(1)$The terms $a_{2i}$ are $a_2, a_4, \dots, a_n$. This is an arithmetic progression with first term $a_2 = a_1 + 1$ and common difference $2d = 2$.The number of terms is $n/2$.$\sum_{i=1}^{n/2} a_{2i} = \frac{n/2}{2}[2(a_1+1) + (n/2 - 1)2] = 120$.$\frac{n}{4}[2a_1 + 2 + n - 2] = 120 \Rightarrow \frac{n}{4}[2a_1 + n] = 120$.$2a_1 + n = \frac{480}{n} \quad \dots(2)$Subtracting $(1)$ from $(2)$:$(2a_1 + n) - (2a_1 + n - 1) = \frac{480}{n} - \frac{384}{n}$.$1 = \frac{96}{n}$.$n = 96$.

If $\{a_{i}\}_{i=1}^{n}$,where $n$ is an even integer,is an arithmetic progression with common difference $d=1$,and $\sum_{i=1}^{n} a_{i}=192$,$\sum_{i=1}^{n/2} a_{2i}=120$,then $n$ is equal to

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