If $\sum\limits_{i=1}^{21}a_i= 693$, where $a_1, a_2, ...,a_{21}$ are in AP, then the value of $\sum\limits_{i=0}^{10}a_{2r+1}$ is

363

$∵ a_1, a_2,....a_{21}$, are in AP

$∴ a_1 + a_2 +....+a_{21} =\frac{21}{2}(a_1+a_{21})$

$⇒ 693 =\frac{21}{2}(a_1+a_{21})$   (given)

$∴ a_1 + a_{21}1 = 66$ ....(i)

$∴\sum\limits_{i=0}^{10}a_{2r+1}=a_1 + a_3 + a_5 + a_7 + a_9 +...+a_{21}$

$=(a_1 + a_{21}) + (a_3 + a_{19}) + (a_5 + a_{17}) + (a_7 + a_{15}) + (a_9 + a_{13}) + a_{11}$

$= 5×(a_1 + a_{21}) + a_{11} (∴ T_n + T_n’ = a + l)$

$= 5 × 66 + a_{11}= 330 + a_{11}$

$=330+(\frac{a_1+a_{21}}{2})$  ($∵ a_{11}$ is middle term)

$= 330 + 33 = 363$