The sum of 8th term of an AP is 160 while the sum of 20th is 880 fine the 43 term and the sum of the 12 term
"sum of 8th term" makes no sense. A sum has to add two things.
Now, if you meant to say "sum of the first 8 terms" then you have
8/2 (2a+7d) = 160
20/2 (2a+19d) = 880
Then solve for a and d, then find
a + 42d
and
12/2 (2a+11d)
We're doing math here. If you can't say what you mean, then you have no hope of getting an answer that means anything.
Help me
14
Oh, the world of Arithmetic Progressions, where numbers march in an orderly fashion! Let's dive in and find some answers, shall we?
To find the 43rd term, we can use the sum of the two given terms, namely the 8th and 20th terms. Imagine a train of numbers passing by, and let's call the common difference between them "d".
Now let's set up some equations:
For the 8th term: a + 7d = 160
For the 20th term: a + 19d = 880
Let's solve these equations using some not-so-serious mathematical skills... Drumroll, please!
(A comedian walks on stage) Alright, let's start our imaginary math show! (Imitates suspenseful music) And the 43rd term is... (Wait for it...) (Knock-knock) Honey, can you answer that? (Audience groans) Oops, looks like we got sidetracked! Let's solve the problem properly now.
Solving the equations, we find that the common difference "d" is 20, and the first term "a" is 0. So, the 43rd term is 0 + 42 * 20 = 840. Ta-da!
Now, onto the sum of the 12th term. A touch of mathematics, a dab of humor, and a sprinkle of silliness... Let's go!
The sum of the 12th term can be determined using the formula for the sum of an Arithmetic Progression:
Sn = (n/2)(2a + (n-1)d)
Plugging in the values, we have:
S12 = (12/2)(2a + (12-1)d)
S12 = 6(2a + 11d)
But wait! I'm feeling generous today, so let's add a little extra fun factor. (Puts on a clown wig) To calculate the sum of the 12th term, we need to do some calculation magic: Abracadabra, hocus pocus, and voila! The final answer is... (Enthusiastically) 6 times the sum of the second and ninth terms, raised to the power of the first term! (Audience laughs)
Alright, alright, let's get back to seriousness. The actual answer is:
S12 = 6(2(0) + 11(20))
S12 = 6(0 + 220)
S12 = 6(220)
S12 = 1320.
There you have it! The sum of the 12th term is 1320. Now, go forth and conquer those Arithmetic Progressions, my friend!
To find the 43rd term and the sum of the 12th term of an arithmetic progression (AP), we need to use the formulas related to the arithmetic progression.
The formula to find the nth term of an AP is given by:
an = a + (n-1)d
where,
an is the nth term,
a is the first term,
n is the term number,
d is the common difference between the consecutive terms.
We are given that the sum of the 8th term is 160. Using this information, we can determine the value of the 8th term.
Let's substitute the given values into the formula:
160 = a + (8-1)d
160 = a + 7d
Similarly, we are given that the sum of the 20th term is 880. Using this information, we can determine the value of the 20th term.
Substituting the given values into the formula, we get:
880 = a + (20-1)d
880 = a + 19d
Now, we have two equations with two variables (a and d):
160 = a + 7d
880 = a + 19d
We can solve these two equations simultaneously to find the values of a and d.
Let's subtract the first equation from the second equation to eliminate 'a':
880 - 160 = (a + 19d) - (a + 7d)
720 = 19d - 7d
720 = 12d
To find the value of 'd', we need to divide both sides of the equation by 12:
d = 720/12
d = 60
Now, substitute the value of 'd' back into the first equation to find the value of 'a':
160 = a + 7(60)
160 = a + 420
a = 160 - 420
a = -260
So, the first term (a) is -260, and the common difference (d) is 60.
To find the 43rd term, substitute the values of 'a', 'd', and 'n' into the formula:
a43 = -260 + (43-1)60
a43 = -260 + 42*60
a43 = -260 + 2520
a43 = 2260
Therefore, the 43rd term of the arithmetic progression is 2260.
To find the sum of the 12 terms, we can use the formula for the sum of an arithmetic series:
Sn = (n/2)(2a + (n-1)d)
Substituting the given values into this formula:
S12 = (12/2)(2*(-260) + (12-1)*60)
S12 = 6*(-520 + 11*60)
S12 = 6*(-520 + 660)
S12 = 6*140
S12 = 840
So, the sum of the first 12 terms of the arithmetic progression is 840.