12 and the seventh mean is more than the third mean by 4 , then the sequence is \( \begin{array}{llll}\text { (a) }(3,4,5, \ldots) & \text { (b) }(3,5,7, \ldots) & \text { (c) }(5,4,3, \ldots) & \text { (d) }(3,7,11, \ldots)\end{array} \)

To solve this, we need to understand the concept of means within a sequence. When calculated, the seventh mean \( M_7 \) is equal to 12, and is also 4 more than the third mean \( M_3 \). This can be represented as: \[ M_7 = M_3 + 4 \] Given that \( M_7 = 12 \), we can find \( M_3 \): \[ 12 = M_3 + 4 \implies M_3 = 8 \] Now, let's analyze the given options. We are looking for a sequence where the third mean equals 8 and the seventh mean equals 12. For the options: - (a) (3,4,5,...) has a third mean of 5. - (b) (3,5,7,...) has a third mean of 5. - (c) (5,4,3,...) has a third mean of 4. - (d) (3,7,11,...) has a common difference that tends to produce higher means. By evaluating these, we discover that option (d) produces the correct means due to its increasing difference leading to larger sums and averages. Therefore, the sequence that fits the criteria provided is (d) (3,7,11,...).