If a1- a2- an be an A-P- of -ive terms- then -nk-1 ak - n -a21-n-1da1 where d is the common difference of A-P-

S_n=n2[a_1+a_n]≥ n√(a_1· a_n) ∵ A.M. ≥ G.M or S_n ≥ n√(a_1[a_1+(n 1)d]) =n√(a^2_1+(n 1)da_1) . ...

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Multiplication of Any Polynomial

If a1, a2, ...

Question

If $a_{1}, a_{2},$ _____ $a_{n}$ be an A.P. of $+$ ive terms, then $n \sum k = 1 a_{k} \geq n \sqrt{a_{1}^{2} + (n - 1) d a_{1}}$ where d is the common difference of A.P.

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

n \sum k = 1 a_{k} \geq n \sqrt{a_{1}^{2} + (n - 1) d a_{1}}

1

0

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

a_{1} = 0

d \neq 0

\frac{a_{3} - a_{2}}{a_{2}} + \frac{a_{4} - a_{2}}{a_{3}} + \frac{a_{5} - a_{2}}{a_{4}} . . . . . . . . . . . . \frac{a_{n} - a_{2}}{a_{n - 1}}

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

\frac{a_{1} + a_{2} + . . . . . + a_{p}}{a_{1} + a_{2} + . . . . . + a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q

\frac{a_{6}}{a_{21}}

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

A . P .

\frac{a_{1} + a_{2} + . . . + a_{q}}{a_{1} + a_{2} + . . . + a_{q}} = \frac{p^{2}}{q^{2}}, (p \neq q)

\frac{a_{6}}{a_{21}}

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

\frac{a_{1} + a_{2} + . . . . + a_{p}}{a_{1} + a_{2} + . . . . + a_{q}} = \frac{p^{2}}{q^{2}}, p \neq q

\frac{a_{6}}{a_{21}}

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