Question

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 fifteenth terms of the A.P.

Answer 1

Let a be the first term and d be the common difference of the given AP.

Given: S9 = 162  & a6 : a13 = 1 : 2

S9 = 9/2[2a +(9-1)d]
[ Sn = n/2 [2a + (n-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
a + (6-1)d : a +(13-1)d= 1:2
[an = a +(n-1)d]

a + 5d / a +12d = ½
2( a +5d ) = a +12d

2a + 10d = a +12d
2a - a = 12d -10d
a = 2d……………………..(2)

Put the value of a  from eq 2 in eq 1
18 = a +4d
18 = 2d +4d
18 = 6d
d = 18/6
d = 3

Put the value of d in eq 2
a = 2d
a = 2(3)
a = 6

First term = 6

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 2

Answer:

Step-by-step explanation:see the answer below