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The sum of first four terms of a G.P. is 160 and the common ratio is 3 , then the $4^{\text {th }}$
term is
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term is
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Verified Answer
The correct answer is:
108
$a, a r, a r^{2}, a r^{3}$
$2=3$
$\frac{a\left(\varepsilon^{4}-1\right)}{(\varepsilon-1)}=160$
Solving these we get $a=4$
$a \varepsilon^{3}=t_{4}=108$
$2=3$
$\frac{a\left(\varepsilon^{4}-1\right)}{(\varepsilon-1)}=160$
Solving these we get $a=4$
$a \varepsilon^{3}=t_{4}=108$
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