The sum of first q terms of an AP is 162. the ratio of 6th term and 13...
Given,
Sum of terms= 162
Ratio of 6th term and 13th term= 1:2
Formula is an= a1+(n-1)d
Then the sixth term a6= a1+5d
The 13th term is a13=a1+12d
As given a6/a13 is 1/2
(a1+5d)/(a1+12d)=1/2
After cross multiplication
2(a1+5d)=1(a1+12d)
2a1+10d=a1+12d
2a1-a1=12d-10d
Therefore the first term a1=2d
Sn=(n/2)[2a+(n-1)d]
S9=(9/2)[2(2d)+(9-1)d]
S9=(9/2)(4d+8d)
162=(9/2)(12d)
162=9(6d)
54d=162
d=3
a1=2d=2(3)=6
a15=6+(15-1)3=6+3(14)=48
The first term of A.P is 6
The 15th term of A.P is 48
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The sum of first q terms of an AP is 162. the ratio of 6th term and 13...
The sum of first q terms of an AP is 162. the ratio of 6th term and 13...
Given:
- Sum of the first q terms of an AP is 162.
- The ratio of the 6th term and the 13th term is 1:2.
To Find:
- First term (a) and 15th term (l) of the AP.
Explanation:
1. Sum of the first q terms of an AP:
The sum of the first q terms of an arithmetic progression (AP) is given by the formula:
Sq = (q/2)(2a + (q-1)d)
In this case, the sum of the first q terms is given as 162. So, we have:
162 = (q/2)(2a + (q-1)d) ...........(1)
2. Ratio of the 6th term and 13th term:
The formula to find the nth term (Tn) of an AP is:
Tn = a + (n-1)d
Given that the ratio of the 6th term (T6) and the 13th term (T13) is 1:2, we can write:
T6/T13 = 1/2
Using the formula for the nth term, we can substitute the values:
(a + 5d)/(a + 12d) = 1/2 ...........(2)
3. Solving the Equations:
To find the first term (a) and the common difference (d), we need to solve equations (1) and (2) simultaneously.
From equation (2), we can rewrite it as:
2(a + 5d) = a + 12d
2a + 10d = a + 12d
a = 2d ...........(3)
Substituting the value of a from equation (3) into equation (1), we get:
162 = (q/2)(4d + (q-1)d)
324 = q(4d + (q-1)d)
324 = q(5d + (q-1)d)
324 = q(5d + qd - d)
324 = q(6qd - d)
324 = q(6qd - d)
324 = 6qd^2 - qd
6qd^2 - qd - 324 = 0
Now, we can solve this quadratic equation to find the values of q and d. Once we have q and d, we can substitute them into equation (3) to find the first term (a).
4. Finding the 15th term:
Once we have the values of a and d, we can use the formula for the nth term to find the 15th term (T15):
T15 = a + (15-1)d
5. Conclusion:
By solving the equations and substituting the values, we can find the first term (a), the 15th term (T15), and the common difference (d) of the arithmetic progression (AP).
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