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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?
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The sum of first q term of an A.P is 162. The ratio of 6th term to it'...
It think the question might be 9 terms instead of q terms. if it is q terms then we would not get the values of the terms in numbers.๐Ÿ˜
Here is the solution of the problem if the sum of 9 terms of A.P is 162.๐Ÿ˜Š
Given,
Sum of terms= 162๐Ÿ˜
Ratio of 6th term and 13th term= 1:2๐Ÿ˜…
Formula is an= a1+(n-1)d
Then the sixth term a6= a1+5d๐Ÿ”ข
The 13th term is a13=a1+12d๐Ÿ‘‘
As given a6/a13 is 1/2๐Ÿ˜•
(a1+5d)/(a1+12d)=1/2๐Ÿ‘
After cross multiplication
2(a1+5d)=1(a1+12d)๐Ÿ˜•
2a1+10d=a1+12d๐Ÿ˜๏ธ
2a1-a1=12d-10d๐ŸŽ‡
Therefore the first term a1=2d๐Ÿ˜ต
Sn=(n/2)[2a+(n-1)d]๐Ÿ˜ž
S9=(9/2)[2(2d)+(9-1)d]๐Ÿ˜ค
S9=(9/2)(4d+8d)๐Ÿ˜–
162=(9/2)(12d)๐ŸฆŠ
162=9(6d)๐Ÿถ
54d=162๐Ÿฉ
d=3๐Ÿ˜–
a1=2d=2(3)=6๐Ÿ˜”
a15=6+(15-1)3=6+3(14)=48๐Ÿ˜‹
The first term of A.P is 6๐Ÿ˜œ
The 15th term of A.P is 48๐Ÿ‘Œ๐Ÿ‘Œ
Community Answer
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)
q2
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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?
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