GM for 8 numbers a1- a2- a3-a8 will be given by a1 - a2 - a3 - a88

The correct option is B FalseGM of a_1,a_2 ...... a_8 is given by √(a_1.a_2.a_3.a_4.a_5.a_6.a_7.a_8) Hence the given statement is false. ...

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GM for 8 nu...

Question

GM for $8$ numbers $a_{1}, a_{2}, a_{3} . . . . . a_{8}$ will be given by $(a_{1} \times a_{2} \times a_{3} \times . . . . . . \times a_{8})^{8}$

True

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False

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a_{1}, a_{2}, a_{3}, . . ., a_{15}

a_{1} + a_{8} + a_{15} = 15

a_{2} + a_{3} + a_{8} + a_{13} + a_{14} =

a_{1}, a_{2}, a_{3} . . . . . . ., a_{15} a r e i n A . P a n d a_{1} + a_{8} + a_{15} = 15 t h e n a_{2} + a_{3} + a_{8} + a_{13} + a_{14}

If $a_{1}, a_{2}, \cdot \cdot \cdot, a_{15}$ are in A.P. and $a_{1} + a_{8} + a_{15} = 15$ , then $a_{2} + a_{3} + a_{8} + a_{13} + a_{14}$ equals

A . P .

a_{1}, a_{2}, a_{3}, . . . . . ., a_{n}, . . . . .

a_{1} + a_{3} + a_{5} = - 12

a_{1} a_{2} a_{3} = 8

a_{2} + a_{4} + a_{6}

a_{1}, a_{2}, a_{3}, . . . . . . . . ., a_{n}

(a_{1} + a_{2} + a_{3} + . . . . . . . + a_{n}) (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} . . . . . . . . . . . . + \frac{1}{a_{n}}) \geq n^{2}

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