There are 4 numbers such that first 3 of them form an AP and the last 3 of them form a GP. The sum of the first and third number is 2 and that of second and fourth is 26. what are these numbers?
Rainbowcat203:
Wht is GP
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♧♧HERE IS YOUR ANSWER♧♧
Let, the numbers are a, b, c and d.
Given that,
a, b and c form an AP, then
b - a = c - b
=> 2b = a + c .....(i)
and
b, c and d form a GP, then
c/b = d/c
=> c² = bd .....(ii)
Also, sum of a and c is 2, and sum of b and d is 26.
Then,
a + c = 2 .....(iii)
and
b + d = 26 .....(iv)
Using (i) and (iii), we eliminate (a + c) and we get :
2b = 2
=> b = 1
Putting b = 1 in (iv), we get :
1 + d = 26
=> d = 25
Putting b = 1 and d = 25 in (ii), we get :
c² = 1 × 25 = 25 = 5²
=> c = ± 5
Putting b = 1 and c = 5 in (i), we get :
2 × 1 = a + 5
=> a = - 3
Again, putting b = 1 and c = - 5 in (ii), we get :
2 × 1 = a - 5
=> a = 7
Therefore, we have a set of numbers as :
a = - 3, b = 1, c = 5, d = 25
and
a = 7, b = 1, c = - 5, d = 25
So, we can verify AP and GP conditions, given by (i) and (ii) respectively.
♧♧HOPE THIS HELPS YOU♧♧
Let, the numbers are a, b, c and d.
Given that,
a, b and c form an AP, then
b - a = c - b
=> 2b = a + c .....(i)
and
b, c and d form a GP, then
c/b = d/c
=> c² = bd .....(ii)
Also, sum of a and c is 2, and sum of b and d is 26.
Then,
a + c = 2 .....(iii)
and
b + d = 26 .....(iv)
Using (i) and (iii), we eliminate (a + c) and we get :
2b = 2
=> b = 1
Putting b = 1 in (iv), we get :
1 + d = 26
=> d = 25
Putting b = 1 and d = 25 in (ii), we get :
c² = 1 × 25 = 25 = 5²
=> c = ± 5
Putting b = 1 and c = 5 in (i), we get :
2 × 1 = a + 5
=> a = - 3
Again, putting b = 1 and c = - 5 in (ii), we get :
2 × 1 = a - 5
=> a = 7
Therefore, we have a set of numbers as :
a = - 3, b = 1, c = 5, d = 25
and
a = 7, b = 1, c = - 5, d = 25
So, we can verify AP and GP conditions, given by (i) and (ii) respectively.
♧♧HOPE THIS HELPS YOU♧♧
Answered by
2
Solution:
_____________________________________________________________
Given:
Statement 1 : There are 4 numbers ,...
let first number be a,
let second number be b,
let third number be c,
let forth number be d,
remaining part of statement 1 : first 3 numbers form AP and last 3 numbers form GP.
Statement 2 : The first 3 of the 4 numbers form AP,.
The sum of first and third is 2,.
which means,
a + c = 2,
as per AP,
a+(a+2d) = 2a + 2d = 2,.
As we know that first 3 numbers form AP,
so,
second term is a+d,.
∴ = b,.
and hence,.
2a + 2d = 2,
=> a + d = 1,.. b=1,.
___________________
Statement 3 :
b + d = 26
1 + d = 26
=> d = 25
__________________
As per statement 1,
remaining part,. last 3 form GP series,.
b is first term in G.P
and d is 3third & last term in GP series,.
a = 1, ar² = 25,.
r² = 25,.
and so, r =5,.
As r = 5,.
c = br = 1 x 5 = 5,.
________________
As,
first term of the AP series,.
a + c = 2,
a + 5 = 2
a = 2 - 5
a = -3,.
__________
Hence,
The 4 numbers are,.
=> -3,1,5,25..
_____________________________________________________________
Hope it Helps!!
_____________________________________________________________
Given:
Statement 1 : There are 4 numbers ,...
let first number be a,
let second number be b,
let third number be c,
let forth number be d,
remaining part of statement 1 : first 3 numbers form AP and last 3 numbers form GP.
Statement 2 : The first 3 of the 4 numbers form AP,.
The sum of first and third is 2,.
which means,
a + c = 2,
as per AP,
a+(a+2d) = 2a + 2d = 2,.
As we know that first 3 numbers form AP,
so,
second term is a+d,.
∴ = b,.
and hence,.
2a + 2d = 2,
=> a + d = 1,.. b=1,.
___________________
Statement 3 :
b + d = 26
1 + d = 26
=> d = 25
__________________
As per statement 1,
remaining part,. last 3 form GP series,.
b is first term in G.P
and d is 3third & last term in GP series,.
a = 1, ar² = 25,.
r² = 25,.
and so, r =5,.
As r = 5,.
c = br = 1 x 5 = 5,.
________________
As,
first term of the AP series,.
a + c = 2,
a + 5 = 2
a = 2 - 5
a = -3,.
__________
Hence,
The 4 numbers are,.
=> -3,1,5,25..
_____________________________________________________________
Hope it Helps!!
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