If a1, a2, a3 ...... an are in H.P and f(k)=∑r=1n ar−ak then a1f(1),a2f(2),a3f(3)……,anf(n) are in

f(k)+ak=∑r=1n ar⇒a1+f(1)=a2+f(2)=a3+f(3)+…=an+f(n)Given a1, a2 .... an are in H.P⇒a1+f(1)a1,a2+f(2)a2,….,an+f(n)an are in A.P ⇒a1f(1),a2f(2),…………anf(n) are in H.P

If a1, a2, a3 ...... an are in H.P and f(k)=∑r=1n ar−ak then a1f(1),a2f(2),a3f(3)……,anf(n) are in

f(k)+ak=∑r=1n ar⇒a1+f(1)=a2+f(2)=a3+f(3)+…=an+f(n)Given a1, a2 .... an are in H.P⇒a1+f(1)a1,a2+f(2)a2,….,an+f(n)an are in A.P ⇒a1f(1),a2f(2),…………anf(n) are in H.P

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If a₁, a₂, a₃ ...... a_n are in H.P and $f (k)= (\sum_{r = 1}^{n} a_{r}) - a_{k} then \frac{a_{1}}{f (1)}, \frac{a_{2}}{f (2)}, \frac{a_{3}}{f (3)} \dots \dots, \frac{a_{n}}{f (n)}$ are in

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Detailed Solution

$\begin{array}{l} f (k) + a_{k} = \sum_{r = 1}^{n} a_{r} \\ \Rightarrow a_{1} + f (1) = a_{2} + f (2) = a_{3} + f (3) + \dots = a_{n} + f (n) \end{array}$
Given a₁, a₂ .... an are in H.P
$\begin{array}{l} \Rightarrow \frac{a_{1} + f (1)}{a_{1}}, \frac{a_{2} + f (2)}{a_{2}}, \dots ., \frac{a_{n} + f (n)}{a_{n}} are in A . P \\ \Rightarrow \frac{a_{1}}{f (1)}, \frac{a_{2}}{f (2)}, \dots \dots \dots \dots \frac{a_{n}}{f (n)} are in H . P \end{array}$