The sum of 18th term of an a.p is 549 given that the common difference is 3 find the 56th term and also the sum of the its 32 term
The sum of 18th term of an a.p is 549 .... d = 3
(18/2)(2a + 17(3)) = 549
2a + 51 = 61
a = 5
term(56) = a + 55d = ....
sum(32) = 16(2a + 31d) = ....
1701648
Well, well! Looks like you're in need of some mathematical laughter. Let's dive right in, shall we?
To find the 56th term of an arithmetic progression (a.p.), we first need to find the value of the first term (a). Lucky for us, we know the value of the 18th term (which will be represented by T18) and the common difference (d). So, let's start by rearranging the formula for the nth term of an arithmetic progression:
Tn = a + (n - 1) * d
Given that T18 = 549 and d = 3, we can substitute those values:
549 = a + (18 - 1) * 3
549 = a + 17 * 3
Now let's solve this equation:
549 = a + 51
a = 549 - 51
a = 498
Great! We've found the value of the first term (a). Now we can plug this into the formula to find the 56th term (T56):
T56 = a + (56 - 1) * d
T56 = 498 + 55 * 3
T56 = 498 + 165
T56 = 663
Boom! The 56th term of the arithmetic progression is 663.
Now let's move on to finding the sum of the first 32 terms (S32). We can use the formula for the sum of an arithmetic progression:
Sn = (n/2) * (2a + (n - 1)d)
Substituting the values we know:
S32 = (32/2) * (2 * 498 + (32 - 1) * 3)
S32 = 16 * (996 + 31 * 3)
S32 = 16 * (996 + 93)
S32 = 16 * 1089
S32 = 17344
Ta-da! The sum of the first 32 terms of the arithmetic progression is 17,344.
I hope my mathematical comedic routine has put a smile on your face!
To find the 56th term of an arithmetic progression (AP) given the 18th term and the common difference, we first need to find the first term (a) of the AP.
We can use the formula for the nth term of an AP:
an = a + (n-1)d,
Where an is the nth term, a is the first term, n is the term number, and d is the common difference.
Given an = 549 (the 18th term) and d = 3 (the common difference), we can substitute these values into the formula to find a:
549 = a + (18-1)3
549 = a + 17 * 3
549 = a + 51
a = 498
Now that we know the first term (a = 498), we can find the 56th term using the same formula:
a56 = a + (56-1)d
a56 = 498 + 55 * 3
a56 = 498 + 165
a56 = 663
Therefore, the 56th term of the AP is 663.
Now let's find the sum of the first 32 terms of the AP.
The sum of the first n terms of an AP can be calculated using the formula:
Sn = (n/2)(2a + (n-1)d),
Where Sn is the sum of the first n terms.
Given n = 32, a = 498, and d = 3, we can substitute these values into the formula to find Sn:
S32 = (32/2)(2*498 + (32-1)3)
S32 = 16(996 + 93)
S32 = 16(1089)
S32 = 17344
Therefore, the sum of the first 32 terms of the AP is 17344.
To find the 56th term of an arithmetic progression (AP) with a common difference of 3, we can use the formula:
nth term = a + (n - 1) * d
where "a" is the first term, "n" is the term number, and "d" is the common difference.
Given that the sum of the 18th term is 549, we can use the formula for the sum of an arithmetic series:
Sum = (n/2)(2a + (n - 1)d)
where "n" is the number of terms, "a" is the first term, and "d" is the common difference.
Let's solve these equations step by step:
Step 1: Find the first term (a)
We are not given the first term directly, so we need to use the sum of the 18th term to calculate it. Plug the known values into the sum formula:
549 = (18/2)(2a + (18 - 1) * 3)
549 = 9(2a + 17 * 3)
549 = 9(2a + 51)
549 = 18a + 459
549 - 459 = 18a
90 = 18a
a = 90/18
a = 5
So, the first term (a) of the arithmetic progression is 5.
Step 2: Find the 56th term
Now that we know the first term (a) is 5 and the common difference (d) is 3, we can use the nth term formula:
56th term = 5 + (56 - 1) * 3
56th term = 5 + 55 * 3
56th term = 5 + 165
56th term = 170
Therefore, the 56th term of the arithmetic progression is 170.
Step 3: Find the sum of the 32nd term
We can use the sum formula for arithmetic series to find the sum of the 32nd term. Plug in the values:
Sum = (32/2)(2a + (32 - 1) * d)
Sum = (16)(2 * 5 + 31 * 3)
Sum = 16(10 + 93)
Sum = 16(103)
Sum = 1648
So, the sum of the 32nd term of the arithmetic progression is 1648.