IF THE SECOND AND FORTY FIFTH TERM OF AN AP IS 10 AND 96 RESPECTIVELY. FIND THE FIRST TERM AND COMMON DIFFERENCE.ALSO FIND THE SUM OF FIRST FIFTEEN TERMS
T45-T2 = 43d = 86
d=2
I expect now you can find a, and of course,
S45 = a + 44/2 (2*10 + 43d)
or
S45 = 45/2 (2a+44d)
To find the first term and common difference of an arithmetic progression (AP) when the second term and the forty-fifth term are given, we can use the formula:
\(a_n = a_1 + (n-1)d\)
where:
\(a_n\) = nth term
\(a_1\) = first term
\(d\) = common difference
\(n\) = number of terms
Given that the second term (\(a_2\)) is 10 and the forty-fifth term (\(a_{45}\)) is 96, we can substitute these values into the formula:
For \(a_2\):
\(10 = a_1 + (2-1)d\)
For \(a_{45}\):
\(96 = a_1 + (45-1)d\)
Let's solve these equations:
1) \(10 = a_1 + d\)
2) \(96 = a_1 + 44d\)
To solve for the first term (\(a_1\)) and common difference (\(d\)), we can subtract equation 1 from equation 2:
\(96 - 10 = (a_1 + 44d) - (a_1 + d)\)
\(86 = 44d\)
Divide both sides by 44:
\(d = 86/44\)
\(d = 1.9545\) (rounded to 4 decimal places)
Substitute the value of \(d\) into equation 1 to find \(a_1\):
\(10 = a_1 + 1.9545\)
\(a_1 = 10 - 1.9545\)
\(a_1 = 8.0455\) (rounded to 4 decimal places)
Therefore, the first term (\(a_1\)) is approximately 8.0455, and the common difference (\(d\)) is approximately 1.9545.
To find the sum of the first fifteen terms (\(S_{15}\)) of the arithmetic progression, we can use the formula:
\(S_n = \frac{n}{2}(2a_1 + (n - 1)d)\)
Substituting the given values:
\(S_{15} = \frac{15}{2}(2(8.0455) + (15 - 1)(1.9545))\)
Calculating this expression will give us the sum of the first fifteen terms (\(S_{15}\).
To find the first term (a) and the common difference (d) of the arithmetic progression (AP), you can use the formula:
n = a + (n-1) * d
where n is the term number.
Given that the second term (n=2) is 10, and the forty-fifth term (n=45) is 96, we can set up two equations:
For n=2:
2 = a + (2-1) * d
2 = a + d
For n=45:
96 = a + (45-1) * d
96 = a + 44d
Now, we have a system of linear equations. We can solve these equations simultaneously to find the values of a and d.
Subtracting the equation for n=2 from the equation for n=45, you get:
94 = 43d
Dividing both sides by 43, you find that:
d = 94/43
Now substitute the value of d back into the equation for n=2:
2 = a + (94/43)
Multiplying through by 43 to get rid of the fraction, you find:
(43 * 2) = (43 * a) + 94
86 = 43a + 94
Subtracting 94 from both sides, you get:
-8 = 43a
Dividing by 43, you find that:
a = -8/43
So, the first term (a) is -8/43 and the common difference (d) is 94/43.
To find the sum of the first fifteen terms of the AP, you can use the formula for the sum of an AP:
Sn = (n/2) * (a + l)
where Sn is the sum of the first n terms, a is the first term, and l is the last term.
The last term can be calculated using the formula for the nth term of an AP:
ln = a + (n-1) * d
For n=15:
ln = -8/43 + (15-1) * (94/43)
ln = -8/43 + 14 * (94/43)
Calculating ln, you find:
ln = 214/43
Now, substitute the values of a, ln, and n into the sum formula:
Sn = (15/2) * (-8/43 + 214/43)
Sn = (15/2) * (206/43)
Calculating Sn, you find:
Sn = 930/43
Therefore, the sum of the first fifteen terms of the AP is 930/43.