Geometric sequence 3

In geometric sequence is a8 = 312500; a11= 39062500; sn=1953124. Calculate the first item a1, quotient q and n - number of members by their sum s_n.

Result

a1 =  4
q =  5
n =  9

Solution:

a8=a1q7=312500 a11=a1q10=39062500  31250039062500=q7q10  0.008=q−3 0.008−0.333333333333=q q=5  a8=a1q7=a1⋅57=312500 a1=31250057=4a_{ 8 } = a_1 q^{ 7} = 312500 \ \\ a_{ 11 } = a_1 q^{ 10} = 39062500 \ \\ \ \\ \dfrac{ 312500 }{ 39062500 } = \dfrac{ q^{ 7} }{ q^{ 10} } \ \\ \ \\ 0.008 = q^{ -3} \ \\ 0.008^{ -0.333333333333 } = q \ \\ q = 5 \ \\ \ \\ a_{ 8 } = a_1 q^{ 7} = a_1 \cdot 5 ^{ 7} = 312500 \ \\ a_1 = \dfrac{ 312500 }{ 5 ^{ 7}} = 4
 sn=1953124=a1qn−1q−1=5n−15−1 1953124=5n−1 1953125=5n ln⁡1953125=nln⁡5  n=ln⁡1953125ln⁡5 n=9 \ \\ s_n = 1953124 = a_1 \dfrac{ q^n - 1 } { q - 1 } = \dfrac{ 5^n - 1 } { 5 - 1 } \ \\ 1953124 = 5^n - 1 \ \\ 1953125 = 5^n \ \\ \ln 1953125 = n \ln 5 \ \\ \ \\ n = \dfrac{ \ln 1953125 }{ \ln 5 } \ \\ n = 9

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