Let a₁, a₂, a₃,......be an A.P. with a₆ = 2. Then the common difference of this A.P., which maximises the product a₁a₄a₅, is :

The sum
$$1 + {{{1^3} + {2^3}} \over {1 + 2}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 2 + 3}} + ...... + {{{1^3} + {2^3} + {3^3} + ... + {{15}^3}} \over {1 + 2 + 3 + ... + 15}}$$$$ - {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right)$$ is equal to :

The sum
$${{3 \times {1^3}} \over {{1^3}}} + {{5 \times ({1^3} + {2^3})} \over {{1^2} + {2^2}}} + {{7 \times \left( {{1^3} + {2^3} + {3^3}} \right)} \over {{1^2} + {2^2} + {3^2}}} + .....$$ upto 10 terms is:

If a₁, a₂, a₃, ............... a_n are in A.P. and a₁ + a₄ + a₇ + ........... + a₁₆ = 114, then a₁ + a₆ + a₁₁ + a₁₆ is equal to :