the sum of a1+a5+a10+a15+a20+a24=225 find the sum of this a1+a2+a3.........a23+a24=? how to solve this question?
mathematics
You already have the sum of 225. Typos? But you could add them.
75a = 225
To find a, divide both sides by 75.
To solve this question, you first need to figure out the pattern in the given series.
The given series is: a1 + a5 + a10 + a15 + a20 + a24 = 225.
Looking at the series, you can observe that the difference between the indices of each term is 5, except for the last two terms where the difference is 4.
So, you can consider this as an arithmetic progression with a common difference of 5, except for the last two terms where the common difference is 4.
Now, you need to find the 24th term (a24) of this arithmetic progression.
The formula to find the nth term of an arithmetic progression is: an = a1 + (n - 1) * d,
where an is the nth term, a1 is the first term, n is the position of the term, and d is the common difference.
Using the formula, you can find:
a24 = a1 + (24 - 1) * d.
Now, you need to find the sum of a1 + a2 + a3 + ... + a23 + a24.
This is also an arithmetic progression, but with a common difference of 1. You can use the sum formula for an arithmetic progression to find the sum. The formula is:
Sn = n/2 * (a1 + an),
where Sn is the sum of n terms, n is the number of terms, a1 is the first term, and an is the last term.
Using the formula, you can find:
sum = 24/2 * (a1 + a24).
To find the value of the sum, you need to solve the equations. Given that the sum of a1 + a5 + a10 + a15 + a20 + a24 is 225, you can substitute these values into the equation and solve for a1 and d. Once you find those values, you can substitute them into the sum formula to find the sum of the entire series a1 + a2 + a3 + ... + a23 + a24.
assuming that all the a's are terms of an arithmetic progression, you have
a1+a5+a10+a15+a20+a24 = 6a+69d = 225
Now you want to find
S12 = 12/2 (2a+23d) = 24a+276d = 4(6a+69d) = 4*225 = 900