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

<p><p>The first term, $ a = 3 $</p></p> <p><p>Common difference, $ d $</p></p> <p>The formula for the sum of the first $ n $ terms of an A.P. is:</p> <p>$ S_n = \frac{n}{2} [2a + (n-1)d] $</p> <p>Given:</p> <p>$ S_4 = \frac{1}{5}(S_8 - S_4) $</p> <p>This implies:</p> <p>$ 5S_4 = S_8 - S_4 \quad \Rightarrow \quad 6S_4 = S_8 $</p> <p>Substituting the sum formulas:</p> <p>$ 6 \cdot \frac{4}{2}[2 \times 3 + (4-1)d] = \frac{8}{2}[2 \times 3 + (8-1)d] $</p> <p>Simplifying:</p> <p>$ 6 \times 2 [6 + 3d] = 4 [6 + 7d] $</p> <p>$ 12(6 + 3d) = 4(6 + 7d) $</p> <p>$ 72 + 36d = 24 + 28d $</p> <p>$ 36d - 28d = 24 - 72 $</p> <p>$ 8d = -48 $</p> <p>$ d = -6 $</p> <p>Now, to find $ S_{20} $:</p> <p>$ S_{20} = \frac{20}{2} [2 \times 3 + (20-1)(-6)] $</p> <p>$ S_{20} = 10 [6 + 19 \times (-6)] $</p> <p>$ S_{20} = 10 [6 - 114] $</p> <p>$ S_{20} = 10 \times (-108) $</p> <p>$ S_{20} = -1080 $</p> <p>Thus, the sum of the first 20 terms is $-1080$.</p>