Question

The 3rd term of geometric sequence is 10 and the 6th term is 80 find 1st term,common ratio and s10

Answer 1

An=a+(n-1)d

a3=10

a+2d=10

a+4d=80

2d=70

d=45

a+3d=10

a+3×45=10

a=10-135

a=-125

s10=n/2(a+a10)

s10=5(-125+ -125+9×45)

s10=5(-125+ -125+405)

s10=5(285)

s10=1425

Answer 2

Answer:

The first term, common ratio and S₁₀ for given G.P. are 2.5, 2, 2557.5 respectively.

Step-by-step explanation:

Given, for a geometric sequence, the 3rd term, T₃ = ar² = 10

The 6th term, T₆ = 80

We know that the nth term in geometric progression is given by:

$T_n = ar^{n-1}$

The common ratio can be calculated as:

$r =\frac{T_n}{T_{n-1}}$

$\frac{T_6}{T_3}=\frac{ar^5}{ar^2}$

$r^3 = \frac{80}{10}$

$r^3=8$

$r=2$

The first term calculated as:

$a (2)^{3-1}=10$

$4a =10$

$a=2.5$

For a G.P. the sum of the first n terms can be calculated as :

If $r \neq 1$ and $r > 1$

then  $S_n =\frac{ a(r^n - 1)}{(r - 1)}$

Substitute the value of first term, common ratio, n = 10 in above formula:

$S_{10}=\frac{2.5\times ((2)^{10}-1)}{(2-1)}$

$S_{10}=2.5\times (1024-1)$

$S_{10} = 2557.5$

Therefore, the first term, common ratio and S₁₀ are 2.5, 2, 2557.5 respectively.

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