Question
Find the $20^{\text {th }}$ term of the AP whose $7^{\text {th }}$ term is 24 less than the $11^{\text {th }}$ term, first term being 12.
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Answer
Let the first term, common difference and number of term of an AP are $a , d$ and n , resoectively, Given that, first term $(a)=12$ Now by condition,
$7^{th} term (T_7) = 11^{th} term (T_{11}) - 24$
[$\because$ $n^{th}$ term of an AP, $T_n = a + (n - 1)d]$
$\Rightarrow a + (7 - 1)d = a + (11 - 1)d - 24$
$\Rightarrow a + 6d = a + 10d - 24$
$\Rightarrow 24 = 4d$
$\Rightarrow d = 6$
$\therefore 20^{th}$ term of AP, $T_{20} = a + (20 - 1)d$
$20^{th}$ term of AP, $T_{20} = 12 + 19 \times 6 = 126$
Hence, the reduired $20^{th}$ term of an AP is 126.
$7^{th} term (T_7) = 11^{th} term (T_{11}) - 24$
[$\because$ $n^{th}$ term of an AP, $T_n = a + (n - 1)d]$
$\Rightarrow a + (7 - 1)d = a + (11 - 1)d - 24$
$\Rightarrow a + 6d = a + 10d - 24$
$\Rightarrow 24 = 4d$
$\Rightarrow d = 6$
$\therefore 20^{th}$ term of AP, $T_{20} = a + (20 - 1)d$
$20^{th}$ term of AP, $T_{20} = 12 + 19 \times 6 = 126$
Hence, the reduired $20^{th}$ term of an AP is 126.
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