find the sum of the ap; 2+6+18+54+...................+1458
To find the sum of an arithmetic progression (AP), we need to know the first term (a), the common difference (d), and the number of terms (n).
In this case, the first term (a) is 2, and we need to find the common difference (d) to evaluate the series until we reach the term 1458.
To determine the common difference, we can divide any term by the previous term. Let's try:
6 ÷ 2 = 3
18 ÷ 6 = 3
54 ÷ 18 = 3
Since the common difference is the same, we can see that d = 3.
Now, we need to find the number of terms (n). We have the first term (a) as 2 and the last term as 1458. We can use the formula to find the nth term of an AP:
nth term = a + (n - 1) * d
By substituting the given values, we can solve for n:
1458 = 2 + (n - 1) * 3
1458 = 2 + 3n - 3
1456 = 3n - 1
1457 = 3n
n = 1457 / 3
n ≈ 485.66
Since the number of terms (n) cannot be a decimal, we take the largest integer less than or equal to it, making n = 485.
Now that we have the first term (a = 2), the common difference (d = 3), and the number of terms (n = 485), we can find the sum of the AP using the formula:
Sum = (n/2) * [2a + (n - 1)d]
Substituting the values:
Sum = (485/2) * [2(2) + (485 - 1)(3)]
Sum = 242.5 * [4 + 1454]
Sum = 242.5 * 1458
Sum ≈ 354,159
Looks like a GP, with a = 2 and r = 3
Which term number is 1458 ?
ar^(n-1) = 1458
2(3^(n-1)) = 1458
3^(n-1) = 729 , but I know 3^6 = 729
so n-1 = 6 and n = 7
so now you want sum(7) = 2(3^7 - 1)/(3-1) = ...