IF THE SUM OF FIRST q TERMS OF AN A.P. IS 162.THE RATIO OF ITS 6th TERM TO ITS 13th TERM IS 1:2. FIND ITS FIRST AND 15th TERM.
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Muskan5785
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. Here is the solution of the problem if the sum of 9 terms of A.P is 162. 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
Answers & Comments
Here is the solution of the problem if the sum of 9 terms of A.P is 162.
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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