The sum of first q term of an A.P is 162. The ratio of 6th term to it'...
Problem Statement:
The sum of first q term of an A.P is 162. The ratio of 6th term to it's 13th term is 1:2. Find the first and 15th terms of the A.P?
Solution:
Let us assume the first term of A.P be 'a' and the common difference be 'd'.
Then, the sum of first q terms of an A.P is given by,
Sq = (q/2)[2a + (q-1)d] .................(1)
Step 1: Relation between 6th and 13th term of the A.P
Given, the ratio of 6th term to it's 13th term is 1:2.
Therefore,
a+5d/a+12d = 1/2
2(a+5d) = a+12d
a = 2d
Step 2: Relation between q and d
Given, the sum of first q term of an A.P is 162.
By substituting 'a' from step 1 in equation (1), we get,
Sq = (q/2)[4d + (q-1)d]
Sq = (q/2)(5d + (q-2)d)
Sq = (q/2)(6d + (q-6)d)
162 = (q/2)(6d + (q-6)d)
162 = (q/2)(q+6)d ..................(2)
Step 3: Solving equations (1) and (2)
We have two equations and two variables. By solving them, we get the values of 'a' and 'd'.
Multiplying equation (2) by 2, we get,
324 = q(q+6)d
162 = q(q+6)d/2
Substituting the value of Sq from equation (1), we get,
162 = (q/2)[2a + (q-1)d]
162 = (q/2)[2(2d) + (q-1)d]
162 = (q/2)[5d + qd - d]
162 = (q/2)[4qd/2 + 4d/2]
Substituting d = 4 in the above equation,
162 = (q/2)[4q + 8]
162 = 2(q/2)(2q+4)
81 = (q/2)(q+2)
q
2