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If a_{1},a_{2},...a_{n} be an A.P. of +ive terms, then _{k=1

Question

If $a_{1}, a_{2}, ... a_{n}$ be an A.P. of +ive terms, then $k = 1 \sum n a_{k} \geq n a_{1}^{2} + (n - 1) d a_{1}$ where d is common difference of A.P. If you think this is true write $1$ otherwise write $0$ ?

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Emily Discussed

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

be an A.P. of +ive terms, then

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

where d is common difference of A.P. If you think this is true write

1

otherwise write

0

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Text solutionVerified

S_{n} = \frac{n}{2} [a_{1} + a_{n}] \geq n a_{1} . a_{n} ∵ A . M . \geq G . M .

S_{n} \geq n a_{1} [a_{1} + (n - 1) d] = n a_{1}^{2} + (n - 1) d a_{1}

Thus answer is

1

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Question Text	If $a_{1}, a_{2}, ... a_{n}$ be an A.P. of +ive terms, then $k = 1 \sum n a_{k} \geq n a_{1}^{2} + (n - 1) d a_{1}$ where d is common difference of A.P. If you think this is true write $1$ otherwise write $0$ ?
Answer Type	Text solution:1
Upvotes	150