If the first term of an A.P. is 3 and the sum | vedclass

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(B) Given the first term $a = 3$. Let $d$ be the common difference. The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$. The sum of the

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$-1080$

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(B) Given the first term $a = 3$. Let $d$ be the common difference. The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$. The sum of the first four terms is $S_4 = \frac{4}{2}[2(3) + (4-1)d] = 2(6 + 3d) = 12 + 6d$. The sum of the next four terms is $S_8 - S_4$. According to the problem,$S_4 = \frac{1}{5}(S_8 - S_4)$. $5S_4 = S_8 - S_4 \Rightarrow 6S_4 = S_8$. $6 	imes [2(6 + 3d)] = \frac{8}{2}[2(3) + (8-1)d]$. $12(6 + 3d) = 4(6 + 7d)$. $3(6 + 3d) = 6 + 7d$. $18 + 9d = 6 + 7d$. $2d = -12 \Rightarrow d = -6$. The sum of the first $20$ terms is $S_{20} = \frac{20}{2}[2(3) + (20-1)(-6)]$. $S_{20} = 10[6 + 19(-6)] = 10[6 - 114] = 10(-108) = -1080$.

If the first term of an $A.P.$ is $3$ and the sum of its first four terms is equal to one-fifth of the sum of the next four terms,then the sum of the first $20$ terms is equal to

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