Question

For a G.P t3=20,t6=160 find S7 ​

Answer 1

Answer:

S7 = 635

Step-by-step explanation:

Let the first term of GP is a and common ration r

The third term of the GP

$t_3=ar^2=20$

And the 6th term of the GP

$t_6=ar^5=160$

Dividing the 6th term by the third term

$\frac{t_6}{t_3} =\frac{160}{20} =8$

or, $\frac{ar^5}{ar^2} =8$

$\implies r^3=2^3$

$\implies r=2$

Thus $a\times 2^2=20$

or, a = 5

Therefore, the sum upto 7 terms

$S_7=\frac{a(r^7-1)}{2-1}$

\implies $S_7=\frac{5(2^7-1)}{1}$

$\implies S_7=5(128-1)$

$\implies S_7=5\times 127$

$\implies S_7=635$

Hope this helps.

Answer 2

Step-by-step explanation:

Let the first term of GP is a and common ration r

The third term of the GP

t_3=ar^2=20t

3

=ar

2

=20

And the 6th term of the GP

t_6=ar^5=160t

6

=ar

5

=160

Dividing the 6th term by the third term

\frac{t_6}{t_3} =\frac{160}{20} =8

t

3

t

6

=

20

160

=8

or, \frac{ar^5}{ar^2} =8

ar

2

ar

5

=8

\implies r^3=2^3⟹r

3

=2

3

\implies r=2⟹r=2

Thus a\times 2^2=20a×2

2

=20

or, a = 5

Therefore, the sum upto 7 terms

S_7=\frac{a(r^7-1)}{2-1}S

7

=

2−1

a(r

7

−1)

\implies S_7=\frac{5(2^7-1)}{1}S

7

=

1

5(2

7

−1)

\implies S_7=5(128-1)⟹S

7

=5(128−1)

\implies S_7=5\times 127⟹S

7

=5×127

\implies S_7=635⟹S

7

=635

Hope this helps.