the sum of the first 53 terms of an arithmetic series is 4 240 while the seventh term is equal to 20. Find the first term on common difference
Sn=4240.....equatn 1
T7=20....equatn 2..
In arithmetic sequnce T7=a+6d therefore ur equation 2 will now be 20=a+6d.....equation 2.make 'a' the subject of the formula and then use TN=a+(n-1)d to solve for d...substitude 20-6d as ur a and ur n is 53 then solve for common difference
I meant Sn=4240 meaning S53=4240 coz the n is 53
I really don't understand this
To find the first term (a) and the common difference (d) of an arithmetic series, we can use the formulas related to the sum (S) and the nth term (aₙ) of the series.
Given:
Sum of the first 53 terms (S₅₃) = 4,240
The 7th term (a₇) = 20
Step 1: Applying the formula for the sum of an arithmetic series
The sum of the first n terms of an arithmetic series is given by the formula:
S = (n/2) * (2a + (n-1)d)
Since we are given the sum (S₅₃) and the number of terms (n = 53), we can solve for a + (n-1)d:
4,240 = (53/2) * (2a + 52d) [equation 1]
Step 2: Applying the formula for the nth term of an arithmetic series
The nth term of an arithmetic series is given by the formula:
aₙ = a + (n-1)d
Since we are given the 7th term (a₇ = 20), we can also write:
20 = a + (7-1)d
20 = a + 6d [equation 2]
Step 3: Solving the equations simultaneously
We will now solve equations 1 and 2 simultaneously to find the values of a and d.
From equation 2, we can rewrite it as:
a = 20 - 6d [equation 3]
Substituting equation 3 into equation 1, we have:
4,240 = (53/2) * (2(20 - 6d) + 52d)
4,240 = (53/2) * (40 - 12d + 52d)
4,240 = (53/2) * (40 + 40d)
4,240 = (53/2) * (40d + 40) [simplifying]
Next, simplify further:
4,240 = (53/2) * 20(2d + 1)
4,240 = 53 * 20(2d + 1)
4,240 = 53 * 40d + 53 * 20
4,240 = 2,120d + 1,060
Now, isolate the value of d:
2,120d = 4,240 - 1,060
2,120d = 3,180
d = 3,180 / 2,120
d = 1.5
Step 4: Calculating the first term (a)
Substituting the value of d back into equation 3:
a = 20 - 6(1.5)
a = 20 - 9
a = 11
Therefore, the first term (a) of the arithmetic series is 11, and the common difference (d) is 1.5.
Sn=4240.....equatn 1
T7=20....equatn 2..
In arithmetic sequnce T7=a+6d therefore ur equation 2 will now be 20=a+6d.....equation 2.make 'a' the subject of the formula and then use TN=a+(n-1)d to solve for d...substitude 20-6d as ur a and ur n is 53 then solve for common difference...Answers,,ur 'a 'should be -1 nd ur 'd 'should be 7/2