Question

Find the 4 successive terms of a
GP of which the second term is
smaller than the first by 35 and
the third term is larger than fourth term by 560.
​

Answer 1

Answer:

Given:

2nd term of the GP is smaller than 1st term by 35

3rd term is larger than 4th term by 560.

To find: 4 successive terms of the GP.

let a be the the first term and r be the common ratio of the GP.

then nth term of GP, $a_n=ar^{n-1}$

According to the Question,

$a-ar=35$

$ar^{3-1}-ar^{4-1}=560$   ⇒ ar² - ar³ = 560

a ( 1 - r ) = 35 ........................ (1)

ar² ( 1 - r ) = 560  ................. (2)

(2) / (1)

$\frac{ar^2(1-r)}{a(1-r)}=\frac{560}{35}$

$r^2=16$

r = ± 4

when r = 4

⇒ a ( 1 - 4 ) = 35  ⇒ a = -35/3

1st term = -35/3

2nd term = -140/3

3rd term = -560/3

4th term = 2240/3

when r = -4

⇒ a ( 1 - (-4) ) = 35  ⇒ a = 35/5 = 7

1st term = 7

2nd term = 28

3rd term = 112

4th term = 448

Answer 2

Step-by-step explanation:

hope the pic helps

......