The number of terms in an A .P. is even ; the | vedclass

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Question

Let $a,d,$ nad $2n $ be the frist term, common difference and total number of terms of an A.P. respectively i.e. $a + \left( {a + d} \right) + \l

Vedclass · Accepted Answer

$8$

Vedclass · Answer

Let $a,d,$ nad $2n $ be the frist term, common difference and total number of terms of an A.P. respectively i.e. $a + \left( {a + d} ight) + \left( {a + 2d} ight) + ... + \left( {a + \left( {2n - 1} ight)d} ight)$

No. of even terms $=n$, of odd terms $=n$

Sum of odd terms:

${S_o} = \frac{n}{2}\left[ {2a + \left( {n - 1} ight)\left( {2d} ight)} ight] = 24\,\,\,\,$

$ \Rightarrow n\left[ {a + \left( {n - 1} ight)d} ight] = 24\,\,\,\,\,\,\,\,\,\,\,\,.....\left( i ight)$

Sum of even terms :

${S_e} = \frac{n}{2}\left[ {2\left( {a + d} ight) + \left( {n - 1} ight)2d} ight] = 30$

$ \Rightarrow n\left[ {a + d + \left( {n - 1} ight)d} ight] = 30\,\,\,\,\,\,\,\,\,\,\,\,.....\left( {ii} ight)$

Subtracting equation $(i)$ from $(ii)$ , we get $nd=6$      ....$(iii)$

Also, given that last term exceeds the first term by $\frac{{21}}{2}$

$a + \left( {2n - 1} ight)d = a + \frac{{21}}{2}$

$2nd - d = \frac{{21}}{2}$

$ \Rightarrow 2 	imes 6 - \frac{{21}}{2} = d$       $(\,\,nd = 6)$

$d = \frac{3}{2}$

Putting value of $d$ in equation $(iii)$

$n = \frac{{6 	imes 2}}{3} = 4$

Total no, of terms $ = 2n = 2 	imes 4 = 8$

The number of terms in an $A .P.$ is even ; the sum of the odd terms in it is $24$ and that the even terms is $30$. If the last term exceeds the first term by $10\frac{1}{2}$ , then the number of terms in the $A.P.$ is

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