Question

Given T3= 20, T7= 320 of a certain GP.Find T8, S10.
someone please help me with this problem.​

Answer 1

Given:

3rd term (a₃) = 20

7th term (a₇) = 320

To find:

8th term and Sum of first 10 terms.

Solution:

a₃ = ar² = 20 ........(1)

a₇ = ar⁶ = 320 ......(2)

Dividing (2) by (1)

ar⁶ = 320

ar² = 20

---------------

r⁴ = 16

r = 4√16

r = 2

Substituting r in (1)

ar² = 20

a(2)² = 20

4a = 20

a = 5

8th term (a₈) = ar⁷

= 5(2)⁷

= 5 x 128

= 640

$\sf S_n = \frac{a(1-r^{n})}{1-r} \\S_{10} = \frac{5(1-2^{10})}{1-2} \\S_{10} = \frac{5(1-1024)}{-1} \\S_{10} = \frac{5(-1024)}{-1} \\S_{10} = \frac{-5115}{-1}\\S_{10} = 5115$

Sum of first 10 terms is 5115.

Answer 2

As per question,

We have a Geometric progression

with T3 ( Third term ) = 20

and T7 ( Seventh term ) = 320

Let the first term of  GP is 'a' and 'r' be the common ratio between consequent terms.

Third term, T3 = ar²

ar² = 20 ... eqn (ii)

Seventh term, T7 = ar⁶

ar⁶ = 320 ... eqn (iii)

Diving eqn (iii) by (ii),

$\frac{ar^6}{ar^2} =\frac{320}{20}$

$r^4 = 16$

=> r = 2

Using r = 2 in equation (ii),

a(2)² = 20

=> a = 5

We need to find T8 (eight term of GP) and S10 (Sum of first 10 terms of GP)

For T8 (eight term), n = 8

(Using values of a and r here)

T8 = ar⁷ = 5(2)⁷

T8 = 5 × 128

Eight term, T8 = 640

For S10, number of term, n = 10

Since r=2 > 1

Sum of n terms of GP = a(r^n -1) / (r-1)

Sum of 10 terms = 5(2¹⁰ - 1)/(2-1)

S10 = 5 (1023)/1

S10 = 5(1023)

S10, Sum of first 10 terms of GP is 5115