Math, asked by akshaydatta52, 2 months ago

The common ratio of a G.P is 1 and S10=40 then find the first term.​

Answers

Answered by anjaliraut22
2

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

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