Question
If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.
✓
Answer
Given:
$\text{a}_9=0$
$\therefore\text{a}+8\text{d}=0$
$\text{a}=-8\text{d}\ .....(1)$
$\text{a}_{19}=\text{a}+(19-1)\text{d}$
$=\text{a}+18\text{d}$ $[\therefore\text{a}=-8\text{d}\ \text{from}(1)]$
$=10\text{d}\ .....(2)$
$\text{a}_{29}=\text{a}+(29-1)\text{d}$
$=-8\text{d}+28\text{d}$ $[\because\text{a}=-8\text{d}\ \text{from}(1)]$
$-20\text{d}\ .....(3)$
(2) and (3)
$\text{a}_{29}=2\text{a}_{19}$
Hence proved.
$\text{a}_9=0$
$\therefore\text{a}+8\text{d}=0$
$\text{a}=-8\text{d}\ .....(1)$
$\text{a}_{19}=\text{a}+(19-1)\text{d}$
$=\text{a}+18\text{d}$ $[\therefore\text{a}=-8\text{d}\ \text{from}(1)]$
$=10\text{d}\ .....(2)$
$\text{a}_{29}=\text{a}+(29-1)\text{d}$
$=-8\text{d}+28\text{d}$ $[\because\text{a}=-8\text{d}\ \text{from}(1)]$
$-20\text{d}\ .....(3)$
(2) and (3)
$\text{a}_{29}=2\text{a}_{19}$
Hence proved.
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