Let S_n denote the sum of the first n terms of an A.P. If S₄ = 16 and S₆= – 48, then S₁₀ is equal to :

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 :

Let $$a$$, b and c be in G.P. with common ratio r, where $$a$$ $$ \ne $$ 0 and 0 < r $$ \le $$ $${1 \over 2}$$ . If 3$$a$$, 7b and 15c are the first three terms of an A.P., then the 4^th term of this A.P. is :

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 :