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NDA Math Sequences and Series

Consider the following for the next three (03) items that follow:

Let a,x,y,z,b be in A.P,where

x+y+z=15

.Leta,p,q,r,b be in HP, where

p^{-1}+q^{-1}+r^{-1}=\frac{5}{3}

[NDA II 2016]

Question : 60 of 79

Marks: +1, -0

Solution:

S_n = \;\;\frac{n}{2}(a+l)

\Rightarrow a+x+y+z+b = \;\;\frac{5}{2}(a+b)

a+b+15 = \;\;\frac{5}{2}(a+b)

\Rightarrow a+b = 10

...(1)

\;\;\frac{1}{a}+\;\;\frac{1}{p}+\;\;\frac{1}{q}+\;\;\frac{1}{r}+\;\;\frac{1}{b}= \;\;\frac{5}{2}\left(\;\;\frac{1}{a}+\;\;\frac{1}{b}\right)\;\;\frac{1}{a}+\;\;\frac{1}{p}+\;\;\frac{1}{q}+\;\;\frac{1}{r}+\;\;\frac{1}{b}=\;\;\frac{5}{2}\left(\;\;\frac{1}{a}+\;\;\frac{1}{b}\right)

\Rightarrow\;\;\frac{1}{a}+\;\;\frac{1}{b}+\;\;\frac{5}{3}=\;\;\frac{5}{2}\left(\;\;\frac{1}{a}+\;\;\frac{1}{b}\right)

\Rightarrow\;\;\frac{3(a+b)}{ab}=\;\;\frac{10}{3} ....(2)

\Rightarrow\;\;\frac{3\times 10}{ab}=\;\;\frac{10}{3}

\Rightarrow ab=9

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