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If a₂₀ = 4 and a₅ = 10 for an HP, find a₁₀
Option: 1
$\frac{1}{8}$

Option: 2
$\frac{1}{4}$

Option: 3
$\frac{1}{10}$

Option: 4
None of these

Answers (1)

best_answer

Let first term of HP be a₁ and common difference of corresponding AP be d.

So AP is $\frac{1}{a_1},\frac{1}{a_1}+d, \frac{1}{a_1}+2d,......$

Given $a_{20}=4$

$\Rightarrow \frac{1}{a_1}+19d=4$ ........(1)

and $a_5=10$

$\Rightarrow \frac{1}{a_1}+4d=10$ .........(2)

(1) - (2)

$15d=-6$

$d=\frac{-2}{5}$

Using (2)

$\frac{1}{a_1}+4\left (\frac{-2}{5} \right )=10$

$\frac{1}{a_1}=10+\frac{8}{5}=\frac{58}{5}$

So, the 10^th term for AP

$=\frac{1}{a_1}+9d$

$=\frac{58}{5}+9\left (\frac{-2}{5} \right )$

$=\frac{40}{5}=8$

$a_{10}$ for HP $=\frac{1}{8}$

Posted by

Kuldeep Maurya

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