In an A.p ,the 1oth is 68 and the 4th term is 26 .find the first term and the sum of its 100 term
To find the first term and the sum of its 100 terms, we can use the formulas for the nth term and the sum of an arithmetic progression.
Let's assume the first term of the AP is "a" and the common difference is "d".
From the given information, we can write two equations:
a + 9d = 68 (equation 1, as the 10th term is 68)
a + 3d = 26 (equation 2, as the 4th term is 26)
We can solve these two equations simultaneously to find the values of "a" and "d".
Subtracting equation 2 from equation 1, we get:
(a + 9d) - (a + 3d) = 68 - 26
6d = 42
d = 42/6
d = 7
Now substitute the value of d in equation 2:
a + 3(7) = 26
a + 21 = 26
a = 26 - 21
a = 5
Therefore, the first term of the AP is 5.
To find the sum of its 100 terms, we can use the formula:
Sum = (n/2)(2a + (n-1)d)
Here, n = 100, "a" = 5, and "d" = 7.
Sum = (100/2)(2(5) + (100-1)(7))
Sum = 50(10 + 99(7))
Sum = 50(10 + 693)
Sum = 50(703)
Sum = 35150
Therefore, the sum of the 100 terms of the AP is 35150.
To find the first term (a) and the sum of the 100th term (Sn) of an arithmetic progression (AP) given the 10th term (T10) and the 4th term (T4), we can use the following formulas:
Tn = a + (n - 1)d, where Tn represents the nth term.
Sn = (n / 2)(2a + (n - 1)d), where Sn represents the sum of the first n terms.
Given:
T10 = 68
T4 = 26
Step 1: Finding the common difference (d)
Using the formula for Tn, we can find the common difference (d) by setting two equations equal to each other:
T4 = a + (4 - 1)d
T10 = a + (10 - 1)d
26 = a + 3d
68 = a + 9d
Step 2: Solving the two equations simultaneously
To solve these equations simultaneously, we can use a method like substitution or elimination. Let's use the substitution method here:
Using the first equation, we can express a in terms of d:
a = 26 - 3d
Substituting this equation into the second equation:
68 = (26 - 3d) + 9d
68 = 26 + 6d
Step 3: Solving for d
Simplifying the equation:
6d = 68 - 26
6d = 42
d = 7
Step 4: Finding the first term (a)
Now that we have the value of d, we can substitute it back into one of our original equations:
26 = a + 3(7)
26 = a + 21
a = 5
Therefore, the first term (a) of the arithmetic progression is 5.
Step 5: Finding the sum of the 100th term (S100)
Using the formula for Sn, we can find the sum of the first 100 terms:
Sn = (n / 2)(2a + (n - 1)d)
S100 = (100 / 2)(2(5) + (100 - 1)(7))
S100 = 50(10 + 99(7))
S100 = 50(10 + 693)
S100 = 50(703)
S100 = 35,150
Therefore, the sum of the first 100 terms (S100) of the arithmetic progression is 35,150.
To find the first term and the sum of the 100th term of an arithmetic progression (AP) with known terms, we need to use the formulas for term and sum of an AP.
First, let's use the given information to find the common difference (d) of the sequence.
The formula to find the term (an) of an AP is:
an = a + (n - 1)d
Given:
10th term (a10) = 68
4th term (a4) = 26
Using the formula, we can set up two equations:
a10 = a + (10 - 1)d
68 = a + 9d
a4 = a + (4 - 1)d
26 = a + 3d
Now we have a system of equations. We can solve this system of equations simultaneously to find the values of 'a' (first term) and 'd' (common difference).
Subtract the second equation from the first equation to eliminate 'a':
(68 - 26) = (a + 9d) - (a + 3d)
42 = 6d
Now we can substitute the value of 'd' back into one of the original equations to solve for 'a'.
26 = a + 3d
26 = a + 3(7)
26 = a + 21
a = 26 - 21
a = 5
Therefore, the first term (a) of the arithmetic progression is 5.
To find the sum of the first 100 terms of an arithmetic progression, we use the sum formula:
Sn = (n/2)(2a + (n - 1)d)
Given:
First term (a) = 5
Number of terms (n) = 100
Common difference (d) = 7 (as we found earlier)
Substituting these values into the sum formula:
S100 = (100/2)(2 * 5 + (100 - 1) * 7)
S100 = 50(10 + 99 * 7)
S100 = 50(10 + 693)
S100 = 50(703)
S100 = 35,150
Therefore, the sum of the first 100 terms of the arithmetic progression is 35,150.