If a_i_i=1^n, where n is an even integer, is an arithmetic progression with common differe…

Question

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

Accepted Answer

$$\sum\limits_{i = 1}^n {{a_i} = 192} $$

$$\Rightarrow$$ a1 + a2 + a3 + ...... + an = 192

$$ \Rightarrow {n \over 2}[{a_1} + {a_n}] = 192$$

$$ \Rightarrow {a_1} + {a_n} = {{384} \over n}$$ ..... (1)

Now, $$\sum\limits_{i = 1}^{{n \over 2}} {{a_{2i}} = 120} $$

$$\Rightarrow$$ a2 + a4 + a6 + ...... + an = 120

Here total $${n \over 2}$$ terms present.

$$\therefore$$ $${{{n \over 2}} \over 2}[{a_2} + {a_n}] = 120$$

$$ \Rightarrow {n \over 4}[{a_1} + 1 + {a_n}] = 120$$

$$ \Rightarrow {a_1} + {a_n} + 1 = {{480} \over n}$$ ..... (2)

Subtracting (1) from (2), we get

$$1 = {{480} \over n} - {{384} \over n}$$

$$ \Rightarrow 1 = {{96} \over n}$$

$$\Rightarrow$$ n = 96

Solutions · 1