The common ratio of a G.P is 1 and S10=40 then find the first term.
Answers
Letaandrarefirsttermand
commonratioofaG.P.
we now the general term of G.P:
\boxed {\pink { n^{th} \: term = a_{n} = a r^{n-1} }}
n
th
term=a
n
=ar
n−1
4^{th} \: term = 10\: (given)4
th
term=10(given)
\implies a r^{3} = 10 \: ---(1)⟹ar
3
=10−−−(1)
and \: 6^{th} \: term = 40\: (given)and6
th
term=40(given)
\implies a r^{5} = 40 \: ---(2)⟹ar
5
=40−−−(2)
/* Do Equation (2) ÷ equation (1) */
\implies \frac{ar^{5}}{ar^{3}} = \frac{40}{10}⟹
ar
3
ar
5
=
10
40
\implies r^{2} = 4⟹r
2
=4
\implies r = ±\sqrt{2^{2}}⟹r=±
2
2
\implies r = ±2⟹r=±2
/* It is given that common ratio is positive */
Therefore. ,
\implies r = 2 \: ---(3)⟹r=2−−−(3)
/* Substitute r = 2 in the equation (1) , we get */
\implies a\times 2^{3} = 10⟹a×2
3
=10
\implies a = \frac{10}{8}⟹a=
8
10
\implies a = \frac{5}{4} \: ---(4)⟹a=
4
5
−−−(4)
\red { First\: term (a) } \green { = \frac{5}{4}}Firstterm(a)=
4
5
\red { Common \: ratio (r)} \green { = 2}Commonratio(r)=2
Now , a_{n} = 640 \: (given)Now,a
n
=640(given)
\implies ar^{n-1} = 640⟹ar
n−1
=640
\implies \frac{5}{4} \times 2^{n-1} = 640⟹
4
5
×2
n−1
=640
\implies 2^{n-1} = 640 \times \frac{4}{5}⟹2
n−1
=640×
5
4
\implies 2^{n-1} = 128 \times 4⟹2
n−1
=128×4
\implies 2^{n-1} = 2^{9}⟹2
n−1
=2
9
\implies n - 1 = 9⟹n−1=9
\boxed { \pink { Since, If \: a^{m} = a^{n} \implies m = n }}
Since,Ifa
m
=a
n
⟹m=n
\implies n = 10⟹n=10
Therefore.,
\red { First\: term (a) } \green { = \frac{5}{4}}Firstterm(a)=
4
5
\red { Common \: ratio (r)} \green { = 2}Commonratio(r)=2
\red { Number \: of \:terms \: in \:G.P} \green {= 10}NumberoftermsinG.P=10
please check the answer because it get wrong also
