Find $n$ if $1 + 3 + 3^{2} + . . . . . . . + 3^{n - 1} = 364$ .

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Solution

Total number of terms $= n$
First term $= a = 1$
Ratio $= r = 3$

Sum of a GP $= \frac{a (r^{n} - 1)}{r - 1}$

$364 = \frac{1 (3^{n} - 1)}{3 - 1}$

$728 = 3^{n} - 1$

$729 = 3^{n}$

$n = 6$

Hence $n = 6$

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Find n if 1+3+32+.......+3n−1=364.

Total number of terms=nFirst term=a=1Ratio=r=3Sum of a GP=a(rn−1)r−1364=1(3n−1)3−1728=3n−1729=3nn=6Hence n=6

Find $n$ if $1 + 3 + 3^{2} + . . . . . . . + 3^{n - 1} = 364$ .

Total number of terms $= n$
First term $= a = 1$
Ratio $= r = 3$

Sum of a GP $= \frac{a (r^{n} - 1)}{r - 1}$

$364 = \frac{1 (3^{n} - 1)}{3 - 1}$

$728 = 3^{n} - 1$

$729 = 3^{n}$

$n = 6$

Hence $n = 6$