Sequence Series – A.P. – 23 January 2025 (Shift 1)

Solution:

Let first term $a=3$ and common difference be $d$.
Sum of next 4 terms can be written as $S_{next4} = S_8 – S_4$.
Given Condition:
$$S_4 = \frac{1}{5}(S_8 – S_4) \implies 5S_4 = S_8 – S_4 \implies 6S_4 = S_8$$

Using formula $S_n = \frac{n}{2}[2a + (n-1)d]$:
$$6 \cdot \frac{4}{2}[2(3) + 3d] = \frac{8}{2}[2(3) + 7d]$$
$$12(6+3d) = 4(6+7d)$$
$$3(6+3d) = 6+7d$$
$$18 + 9d = 6 + 7d \implies 2d = -12 \implies d = -6$$

Now, calculate $S_{20}$:
$$S_{20} = \frac{20}{2}[2(3) + (20-1)(-6)]$$
$$= 10[6 + 19(-6)]$$
$$= 10[6 – 114]$$
$$= 10(-108) = -1080$$
Ans.(2)