An AP has 20 term and a common difference of 3 determine the last term and the first term of the AP if it's sum is (a)65 (b)650
S20 = 20/2 (2a+19d)
so plug in your numbers and solve for a
Sn=n÷2{2a+19d}
Hmm, let me put on my clown shoes and calculate this for you!
(a) For a sum of 65, I'll need to use a bit of clown magic to figure out the last term. But fear not, for the clown magic is here! By using the formula for the sum of an arithmetic series, which is Sn = (n/2)(2a + (n-1)d), we can plug in the given values. In this case, n = 20, d = 3, and Sn = 65. So, let me do some clown math...
Solving the equation, we get 65 = (20/2)(2a + 19(3)). Simplifying that even further, we have 65 = 10(2a + 57). Divide both sides by 10, and we get 6.5 = 2a + 57. Subtract 57 from both sides, and we end up with -50.5 = 2a. Finally, divide by 2, and we discover that the first term, a, is -25.25. That's quite a clownish number, isn't it?
(b) Now, let's tackle the sum of 650 with the same formula. Using Sn = (n/2)(2a + (n-1)d), we can let n = 20, d = 3, and Sn = 650. Time for some clown calculations!
Solving the equation, we get 650 = (20/2)(2a + 19(3)). Simplifying further, we have 650 = 10(2a + 57). Divide both sides by 10, and we have 65 = 2a + 57. Subtract 57 from both sides, and the result is 8 = 2a. Divide by 2, and we discover that the first term, a, is 4. That's a pretty simple and non-clownish number, isn't it?
So, for (a), the last term of the AP is -25.25 and the first term is -25.25, and for (b), the last term is 4 and the first term is 4. I hope my clownish calculations brought a smile to your face!
To determine the last term and the first term of an arithmetic progression (AP), we need to use the formulas for the nth term and the sum of an AP.
The nth term formula for an AP is given by:
\[ a_n = a_1 + (n - 1)d \]
where \( a_n \) is the nth term, \( a_1 \) is the first term, \( n \) is the number of terms, and \( d \) is the common difference.
The sum of the first \( n \) terms of an AP can be calculated using the formula:
\[ S_n = \frac{n}{2} (2a_1 + (n - 1)d) \]
where \( S_n \) is the sum of the first \( n \) terms.
Now, let's apply these formulas to the given problem:
(a) Sum is 65:
Given that the AP has 20 terms and a common difference of 3, we can substitute these values into the sum formula:
\[ 65 = \frac{20}{2} (2a_1 + (20 - 1)3) \]
Simplifying, we get:
\[ 65 = 10(2a_1 + 19 \cdot 3) \]
\[ 65 = 10(2a_1 + 57) \]
\[ 65 = 20a_1 + 570 \]
\[ 20a_1 = -505 \]
\[ a_1 = -25.25 \]
The first term of the AP is -25.25.
To find the last term, we can use the nth term formula:
\[ a_n = a_1 + (n - 1)d \]
\[ a_{20} = -25.25 + (20 - 1)3 \]
\[ a_{20} = -25.25 + 57 \]
\[ a_{20} = 31.75 \]
Thus, the last term of the AP is 31.75.
(b) Sum is 650:
Applying a similar approach, we can substitute the given values into the sum formula:
\[ 650 = \frac{20}{2} (2a_1 + (20 - 1)3) \]
Simplifying, we have:
\[ 650 = 10(2a_1 + 57) \]
\[ 650 = 20a_1 + 570 \]
\[ 20a_1 = 80 \]
\[ a_1 = 4 \]
The first term of the AP is 4.
Using the nth term formula:
\[ a_{20} = 4 + (20 - 1)3 \]
\[ a_{20} = 4 + 57 \]
\[ a_{20} = 61 \]
So, the last term of the AP is 61.