suppose-that-for-real-numbers-p1p1-p2p2-p15p15-we-have-k115pk30displaystylesumk1

Question

Suppose that for real numbers p1p_{1}, p2p_{2}, ..., p15p_{15} we have ∑k=115pk=30\displaystyle\sum_{k=1}^{15} p_{k} = 30. What is the value of ∑k=115pk6  \displaystyle\sum_{k=1}^{15} \frac{p_{k}}{6}\;?

Accepted Answer

55

Answer

15\frac15

Answer

180180

Answer

52\frac52

Suppose that for real numbers p1p_{1}, p2p_{2}, ..., p15p_{15} we have ∑k=115pk=30\(\displaystyle\)\(\sum\)_{k=1}^{15} p_{k} = 30. What is the value of ∑k=115pk6\(\displaystyle\)\(\sum\)_{k=1}^{15} \(\frac{p_{k}\)}{6}\;?

Suppose that for real numbers $p sub 1$ , $p sub 2$ , ..., $p sub 15$ we have $\sum_{k = 1}^{15} p_{k} = 30$ . What is the value of $\sum_{k = 1}^{15} \frac{p_{k}}{6}$ ?