The first difference of a sequence is 6, 10,14,18. The sum of the first two terms of the original sequence is 24. Find the first three term of the original sequence. the first term of the original sequence is?

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Sra

let the first three terms be a, b, and c

we know
a+b = 24 (#1)
b-a = 6 (#2)
c-b =10 (#3)

add 1 and 2
2b = 30
b=15

in 1 ---> a+15 = 24 ,
a = 9

in 3
c - 15 = 10
c = 25

a=9
b=15
c=25

To solve this problem, let's denote the first term of the original sequence as 'a', and the common difference between each consecutive term as 'd'.

From the given information, we know that the first difference is 6, 10, 14, 18. We can determine the common difference 'd' by subtracting consecutive terms:

10 - 6 = d
14 - 10 = d
18 - 14 = d

Simplifying these equations, we get:
d = 4
d = 4
d = 4

Since the common difference 'd' is the same in all cases, we can conclude that it is constant and equal to 4.

Next, we are given that the sum of the first two terms of the original sequence is 24.

The sum of the first two terms is given by:
a + (a + d) = 24

Substituting the value of 'd' we found earlier:
a + (a + 4) = 24

Simplifying the equation:
2a + 4 = 24
2a = 20
a = 10

Therefore, the first term of the original sequence is 10.

To find the first three terms, we will use the formula for the nth term of an arithmetic sequence:
a + (n - 1) * d

Using 'a' = 10 and 'd' = 4, we can calculate the first three terms of the original sequence as follows:

First term: a = 10
Second term: a + d = 10 + 4 = 14
Third term: a + 2d = 10 + 2 * 4 = 18

So, the first three terms of the original sequence are 10, 14, 18.