# Let $$a_n$$ be the nth term of an AP. If $$\sum_{r=1}^{100}$$ $$a_{2r}$$ = $$\alpha$$ and $$\sum_{r=1}^{100}$$ $$a_{2r-1}$$ = $$\beta$$, then the common difference of the AP is

## Solution :

Given, $$a_2 + a_4 + a_6 + …… + a_{200}$$ = $$\alpha$$      ………(i)

and $$a_1 + a_3 + a_5 + ….. + a_{199}$$ = $$\beta$$           ………(ii)

On subtracting equation (ii) from equation (i), we get

($$a_2 – a_1$$) + ($$a_4 – a_3$$) + ……… + ($$a_{200} – a_{199}$$) = $$\alpha$$ – $$\beta$$

$$\implies$$ d + d + ……..+ 100 times =  $$\alpha$$ – $$\beta$$

$$\implies$$  100d = $$\alpha$$ – $$\beta$$

$$\therefore$$   d = $$\alpha – \beta\over 100$$

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