The sum of 16 terms of an AP is -504,while the sum of its 9 terms is -126.find the sum of its 30 terms
The sum of n terms of an AP:
Sn = ( n / 2 ) ∙ [ 2 a1 + ( n - 1 ) ∙ d ]
The sum of 16 terms of an AP:
S16 = ( 16 / 2 ) ∙ [ 2 a1 + ( 16 - 1 ) ∙ d ]
S16 = 8 ∙ [ 2 a1 + 15 d ] = - 504
8 ∙ 2 a1 + 8 ∙ 15 d = - 504
16 a1 + 120 d = - 504
The sum of 9 terms of an AP:
S9 = ( 9 / 2 ) ∙ [ 2 a1 + ( 9 - 1 ) ∙ d ]
S9 = 4.5 ∙ [ 2 a1 + 8 d ] = - 126
4.5 ∙ 2 a1 + 4.5 ∙ 8 d = - 126
9 a1 + 36 d = - 126
Now you must solve system:
16 a1 + 120 d = - 504
9 a1 + 36 d = - 126
Try it.
The solutions are: a1 = 6 , d = - 5
The sum of 30 terms of an AP :
S30 = ( 30 / 2 ) ∙ [ 2 a1 + ( 30 - 1 ) ∙ d ]
S30 = 15 ∙ [ 2 ∙ 6 + 29 ∙ ( - 5 ) ]
S30 = 15 ∙ ( 12 - 145 ) = 15 ∙ ( - 133 ) = - 1995
I want know how u got the ,6 and-5
Please where's the working for how you got the first term and the common difference??
The 9TH AND THE 22ND term of ap and 29 and 25 respectively. Find the sum of its first 60term
Workin for common difference and first term
Well, it looks like we've got a problem with the arithmetic progression. It must be going through some tough times if the sums are negative! But no worries, I'm here to bring some cheer and solve the problem.
To find the sum of the first 30 terms, we need to first find the common difference (d).
Let's use some clown math! Subtracting the sum of 9 terms (-126) from the sum of 16 terms (-504), we get a difference of -378. Why did the mathematician bring a ladder to the bar? Because he heard the drinks were on the house!
Now, using our calculated difference (-378), we can calculate d. Since we know that the sum of 9 terms equals -126, we can use the formula:
S9 = (9/2) * (2a + 8d) = -126, where a is the first term and d is the common difference.
Re-arranging the formula, we get:
2a + 8d = -28 \*giggles* It's like the math problem is going on a tightrope!
Now, we can plug in the sum of 16 terms (-504) into the sum formula:
S16 = (16/2) * (2a + 15d) = -504.
Re-arranging again, we get:
2a + 15d = -32. Knock, knock. Who's there? D. D who? D best way to solve this math problem!
Now, we have a system of equations:
2a + 8d = -28
2a + 15d = -32.
Subtracting the first equation from the second, we get:
7d = -4.
Solving for d, we find that d is approximately -4/7. Well, in math, fractions are often numb-things. *laughs*
Now that we have d, we can find a using the first equation:
2a + 8d = -28.
Plugging in d, we get:
2a + 8 * (-4/7) = -28.
Simplifying, we find that a is approximately -6. Well, it seems like this arithmetic progression ended up at a circus!
Finally, to find the sum of 30 terms (S30), we use the sum formula:
S30 = (30/2) * (2a + 29d).
Plugging in the values of a and d, we get:
S30 = (30/2) * (2 * (-6) + 29 * (-4/7)).
Calculating that, we find that the sum of the first 30 terms is approximately -365. But hey, don't worry! Even though the sum may be negative, we can always find the positive side of the answer by laughing and enjoying a good joke! *throws confetti*
To find the sum of the 30 terms of an arithmetic progression (AP), we first need to find the common difference (d) and the first term (a₁) of the AP.
Given:
Sum of 16 terms of the AP = -504
Sum of 9 terms of the AP = -126
Using the formula for the sum of the terms of an AP:
Sₙ = (n/2)[2a₁ + (n-1)d]
For the sum of 16 terms:
-504 = (16/2)(2a₁ + (16-1)d)
-504 = 8(2a₁ + 15d) ---(1)
And for the sum of 9 terms:
-126 = (9/2)(2a₁ + (9-1)d)
-126 = 4.5(2a₁ + 8d) ---(2)
To simplify, let's multiply equation (2) by 2:
-252 = 9(2a₁ + 8d) ---(3)
Now, we have a system of equations with equations (1) and (3). We can solve this system of equations using substitution or elimination methods.
Subtract equation (1) from equation (3):
-252 - (-504) = 9(2a₁ + 8d) - 8(2a₁ + 15d)
252 = 9(2a₁ + 8d) - 16a₁ - 120d
252 = 18a₁ + 72d - 16a₁ - 120d
252 = 2a₁ - 48d
Now, we rearrange the equation:
2a₁ - 48d = 252
2(a₁ - 24d) = 252
a₁ - 24d = 126 ---(4)
Now, we can substitute the value of a₁ - 24d from equation (4) into equation (1):
-504 = 8(2a₁ + 15d)
-504 = 16a₁ + 120d
Rearrange the equation:
16a₁ + 120d = -504
Substitute a₁ - 24d from equation (4):
16(a₁ - 24d) + 120d = -504
16a₁ - 384d + 120d = -504
16a₁ - 264d = -504 ---(5)
Now, we have two equations (4) and (5) with two variables (a₁ and d). We can solve these two equations to find out the values of a₁ and d.
Multiply equation (4) by 16:
16(a₁ - 24d) = 16(126)
16a₁ - 384d = 2016 ---(6)
Now, subtract equation (6) from equation (5):
16a₁ - 264d - (16a₁ - 384d) = -504 - 2016
-264d + 384d = -2520
120d = -2520
d = -2520/120
d = -21
Substitute the value of d = -21 into equation (4):
a₁ - 24d = 126
a₁ - 24(-21) = 126
a₁ + 504 = 126
a₁ = 126 - 504
a₁ = -378
Now, we have found the value of d and a₁. We can substitute these values into the sum of 30 terms formula and find the sum of the 30 terms (S₃₀).
S₃₀ = (30/2)[2(-378) + (30-1)(-21)]
S₃₀ = 15[-756 - 29(21)]
S₃₀ = 15[-756 - 609]
S₃₀ = 15(-1365)
S₃₀ = -20,475
Therefore, the sum of the 30 terms of the AP is -20,475.