The first and last terms of an A.p are 10 and 90,if the sum of terms is 750,the number of terms is used to find the common difference if the sum is 1400?
You are telling me that a = 10 and l = 90
sum of tems = (n/2)(a+l) = 750
(n/2)(10+90) = 750
50n = 750
n = 15
Also we know
a + (n-1)d is the last term
10 + 14d = 90
14d = 80
d = 40/7
"the number of terms is used to find the common difference if the sum is 1400?" --- I don't know what that means, but if sum of n terms is 1400
(n/2)(10+90) = 1400
50n = 1400
n= 28
Well, well, well, we're talking about arithmetic progressions (A.P.s) now! Let's do some math and clown around a bit!
To find the common difference (d) when the sum of terms (S) is 1400, we first need to determine the number of terms (n). So let's solve the mystery of n using the given information.
The first term (a₁) is 10, the last term (aₙ) is 90, and the sum of terms (S) is 750. To find n, we'll plug these numbers into the formula for the sum of an A.P.:
S = (n/2)(a₁ + aₙ)
750 = (n/2)(10 + 90)
750 = (n/2)(100)
150 = n/2
n = 300
Ta-dah! We've discovered that there are 300 terms in the A.P. with a sum of 750.
Now, we can move on to the next part of the riddle. The sum of terms (S) is now 1400, and we need to find the common difference (d) with this new sum.
Just like before, we'll use the formula for the sum of an A.P. to solve for d:
S = (n/2)(a₁ + aₙ)
1400 = (300/2)(10 + aₙ)
We already know that a₁ = 10, so let's find aₙ:
aₙ = a₁ + (n - 1)d
aₙ = 10 + (300 - 1)d
aₙ = 10 + 299d
Now we can substitute this into our previous equation:
1400 = (150)(10 + 10 + 299d)
1400 = (150)(20 + 299d)
1400 = 3000 + 44750d
44750d = -1600
d = -1600/44750
d ≈ -0.0357
Well, aren't we feeling a bit negative today! The common difference (d) when the sum of terms is 1400 is approximately equal to -0.0357. Don't let it bring you down, though, I'm here to lighten the mood!
To find the common difference in an arithmetic progression (A.P.), we can use the formula for the sum of an A.P. and the given information.
The formula for the sum of an A.P. is:
sum = (n/2) * (first term + last term)
Given information:
- First term (a₁) = 10
- Last term (aₙ) = 90
- Sum of terms (sum) = 750
Using the sum formula, we can solve for the number of terms (n):
750 = (n/2) * (10 + 90)
750 = (n/2) * 100
7.5 = (n/2) * 10
7.5 = 5n
n = 7.5/5
n = 1.5
Since the number of terms cannot be a fraction, we can round it to the nearest whole number. Therefore, n = 2.
Now, to find the common difference (d) when the sum is 1400, we can use the same formula:
sum = (n/2) * (first term + last term)
Given information:
- First term (a₁) = 10
- Last term (aₙ) = 90
- Sum of terms (sum) = 1400
- Number of terms (n) = 2
Using the sum formula, we can solve for the common difference (d):
1400 = (2/2) * (10 + 90)
1400 = (1) * 100
d = 1400/100
d = 14
Therefore, the common difference in the A.P. when the sum is 1400 is 14.
To solve this problem, we need to use the formula for the sum of an arithmetic progression (A.P.):
Sum of terms (S) = (n/2)(first term + last term)
Given that the first term is 10, the last term is 90, and the sum of terms is 750, we can substitute these values into the formula:
750 = (n/2)(10 + 90)
Now, let's solve for n:
750 = (n/2)(100)
Simplifying further:
750 = 50n
Divide both sides by 50 to isolate n:
n = 750/50
n = 15
Therefore, there are 15 terms in the arithmetic progression.
Now, we need to find the common difference when the sum is 1400.
Using the same formula for the sum of an arithmetic progression, we can solve for the common difference (d):
1400 = (n/2)(10 + 90)
Substituting n = 15:
1400 = (15/2)(100)
Simplifying further:
1400 = 750d
Divide both sides by 750 to isolate d:
d = 1400/750
d = 1.8667
So, when the sum is 1400, the common difference (d) is approximately 1.8667.