The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1 : 2. Find the first and 15th term of the A.P.
Answers
Answer:
The first term is 6 and 15th term is 48.
Step-by-step explanation:
Given :
S9 = 162 & a6 : a13 = 1 : 2
By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]
S9 = 9/2[2a +(9-1)d]
162 = 9/2 [2a +8d]
162 ×2/9 = 2[a +4d]
(18 ×2)/2 = a +4d
18 = a + 4d………………….(1)
a6 : a13 = 1 : 2
By using the formula ,an = a + (n -1)d
a + (6-1)d : a +(13-1)d= 1:2
a + 5d / a +12d = ½
2( a +5d ) = a +12d
2a + 10d = a +12d
2a - a = 12d -10d
a = 2d……………………..(2)
On putting the value of a from eq 2 in eq 1
18 = a +4d
18 = 2d +4d
18 = 6d
d = 18/6
d = 3
On putting the value of d in eq 2
a = 2d
a = 2(3)
a = 6
First term = 6
15th term of the A.P :
a15 = a + (15 - 1)d
a15 = a + 14d
a15 = 6 + 14(3)
a15 = 6 + 42= 48
a15 = 48
Hence, the first term is 6 and 15th term is 48.
HOPE THIS WILL HELP YOU....
Answer:
S9=162
6th term=a+5d
13th term=a+12d
2 (a+5d)=a+12d
2a+10d=a+12d
2a-a=12d-10d
a=2d .....1
S9=162
n/2 (a+l)=162
9/2 (a+a+8d)=162
9/2 (2a+8d)=162
using .....1
9/2 (2×2d+8d)=162
9/2 (4d+8d)=162
9/2 ×12d=162
9×6d=162
54d=162
d=162/54
d=3 ....2
using .....2 in .....1
a=2×3
a=6
a15=a+14d
a15=6+14×3
a15=6+42
a15=48