Question

Question 21 Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

Class X1 - Maths -Sequences and Series Page 193

Answer 1

Let the GP is ar, ar², ar³,......
Given,
3rd term = 1st term + 9
$T_3=T_1+9\\ar^{3-1}=ar^{1-1}+9\\ar^2=a+9\\ar^2-a=9\\\\again,\\2nd-term=4th-term+18\\T_2=T_4+18\\ar^{2-1}=ar^{4-1}+18\\ar=ar^3+18$

now, from both equations ,
$\frac{ar^2-a}{ar-ar^3}=\frac{9}{18}\\\frac{a(r^2-1)}{-ar(r^2-1)}=\frac{1}{2}\\r=-2$

Now, put value of r = -2
a(4) -a = 9
a = 3
Hence,
3, 3(-2), 3(-2)², .....
3, -6 , 12 , -24 , ......

Answer 2

Answer:

Step-by-step explanation: