The first term of an AP is 6 and the fifth term is 18. Find the number of terms in the series having a sum of 162
Let the common difference be d.
Then the second term is 6+d, the third term is 6+2d, the fourth term is 6+3d, and the fifth term is 6+4d.
So we have the equation 6 + 4d = 18.
Solving for d, we find that d = 3.
Now we can write out the terms of the AP: 6, 9, 12, 15, 18, ...
The sum of the first n terms of an AP is given by the formula: S_n = (n/2)(a_1 + a_n), where a_1 is the first term and a_n is the nth term.
We want to find the number of terms, n, such that S_n = 162.
Plugging in the values for a_1 and a_n, we have: 162 = (n/2)(6 + a_n).
We can find a_n by substituting n into the nth term formula: a_n = a_1 + (n-1)d.
Plugging in the values for a_1 and d, we have: a_n = 6 + (n-1)3 = 6 + 3n - 3 = 3n + 3.
Now we can substitute this into the equation for S_n: 162 = (n/2)(6 + 3n + 3).
Simplifying, we get: 162 = (n/2)(3n + 9).
Dividing both sides by 3, we have: 54 = (n/2)(n + 3).
Expanding the equation, we get: 54 = (n^2 + 3n)/2.
Multiplying both sides by 2, we have: 108 = n^2 + 3n.
Rearranging the equation, we get: n^2 + 3n - 108 = 0.
Factoring the quadratic, we have: (n - 9)(n + 12) = 0.
So we have two possible solutions: n = 9 and n = -12.
Since the number of terms cannot be negative, the number of terms is 9.
To find the number of terms in the series, we first need to find the common difference of the arithmetic progression (AP).
We can use the formula for the nth term of an AP:
An = A1 + (n-1)d
Given that the first term (A1) is 6 and the fifth term (A5) is 18, we can substitute these values into the formula:
18 = 6 + (5-1)d
Simplifying the equation, we have:
18 = 6 + 4d
Subtracting 6 from both sides:
12 = 4d
Dividing both sides by 4:
d = 3
Now that we know the common difference (d), we can find the number of terms for a sum of 162.
We can use the formula for the sum of an AP:
Sn = (n/2) * (2A1 + (n-1)d)
Given that A1 = 6, Sn = 162, and d = 3, we can substitute these values into the formula:
162 = (n/2) * (2*6 + (n-1)3)
Simplifying the equation, we have:
162 = (n/2) * (12 + 3n - 3)
Expanding and simplifying further:
162 = (n/2) * (9 + 3n)
Multiplying both sides by 2 to eliminate the fraction:
324 = n(9 + 3n)
Rearranging the equation:
3n^2 + 9n - 324 = 0
Factoring the quadratic equation:
(n + 18)(3n - 18) = 0
Setting each factor equal to zero and solving for n:
n + 18 = 0 or 3n - 18 = 0
n = -18 or n = 6
We discard the negative value since the number of terms cannot be negative, so the number of terms in the series is 6.