Question

Question 14 The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Class X1 - Maths -Sequences and Series Page 192

Answer 1

Given,
sum of first three terms of GP = 16
a + ar +ar² = 16 ---------(1)
again,
sum of next three terms of GP = 128
ar³ + ar⁴ + ar⁵ = 128--------(2)

divide equations (1) and (2),
$\frac{a+ar+ar^2}{ar^3+ar^4+ar^5}=\frac{16}{128}\\\\\frac{a(1+r+r^2)}{ar^3(1+r+r^2)}=\frac{1}{8}\\\\\frac{1}{r^3}=\frac{1}{8}\\r^3=(2)^3\\r=2\\\\\\now\:put\:r=2\\a(1+2+2^2)=16\\a=\frac{16}{7}$
$Now,\:\:S_n=\frac{a(r^n-1)}{r-1}\\\\=\frac{\frac{16}{7}(2^n-1)}{2-1}\\\\=\frac{16}{7}(2^n-1)$

Answer 2

Answer:

$S_n=\frac{16}{7})((2)^{n}-1)$

Step-by-step explanation:

Given : The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128.

To find : Determine the first term, the common ratio and the sum to n terms of the G.P ?

Solution :

The G.P series is $a,ar,ar^2,ar^3,ar^4,ar^5,....$

The sum of first three terms of a G.P. is 16.

i.e. $a+ar+ar^2=16$ ....(1)

The sum of the next three terms is 128.

i.e. $ar^3+ar^4+ar^5=128$ ....(2)

Divide equation (1) and (2),

$\frac{a+ar+ar^2}{ar^3+ar^4+ar^5}=\frac{16}{128}$

$\frac{1+r+r^2}{r^3(1+r+r^2)}=\frac{1}{8}$

$\frac{1}{r^3}=\frac{1}{2^3}$

$r^3=2^3$

$r=2$

The common ratio is r=2.

Substitute in equation (1),

$a(1+r+r^2)=16$

$a(1+2+2^2)=16$

$7a=16$

$a=\frac{16}{7}$

The first term is $a=\frac{16}{7}$

The sum to n terms of the G.P is

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

Substitute the value in the formula,

$S_n=\frac{(\frac{16}{7})((2)^{n}-1)}{(\frac{16}{7})-1}$

$S_n=\frac{(\frac{16}{7})((2)^{n}-1)}{2-1}$

$S_n=\frac{16}{7})((2)^{n}-1)$