Six numbers are in an AP such that their sum | vedclass

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(D) Let the six terms of the $AP$ be $a-5d, a-3d, a-d, a+d, a+3d, a+5d$ with common difference $2d$. Sum of the terms $= (a-5d) + (a-3d) + (a-d) + (a+

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-$4$

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(D) Let the six terms of the $AP$ be $a-5d, a-3d, a-d, a+d, a+3d, a+5d$ with common difference $2d$. Sum of the terms $= (a-5d) + (a-3d) + (a-d) + (a+d) + (a+3d) + (a+5d) = 6a$. Given $6a = 3$,so $a = \frac{1}{2}$. Given $T_1 = 4T_3$,where $T_1 = a-5d$ and $T_3 = a-d$. $a-5d = 4(a-d)$ $\Rightarrow a-5d = 4a-4d$ $\Rightarrow -3a = d$. Substituting $a = \frac{1}{2}$,we get $d = -3 	imes \frac{1}{2} = -\frac{3}{2}$. The fifth term is $T_5 = a+3d$. $T_5 = \frac{1}{2} + 3(-\frac{3}{2}) = \frac{1}{2} - \frac{9}{2} = -\frac{8}{2} = -4$.

Six numbers are in an $AP$ such that their sum is $3$. The first term is $4$ times the third term. Then,the fifth term is:

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