Question

The sum of first 9 terms of an AP is 162. The ratio of its 6th term to its 13th term is 1:2. Find its 1st term

Answer 1

Let a be the first term and d be the common difference.Now, a6a13 = 12⇒a+5da+12d = 12⇒2a+ 10d = a+12d⇒a= 2d    .....(1)Now, sum of first n terms of AP isSn = n2[2a + (n−1)d]⇒S9 = 92[2a + 8d]⇒162 = 9(a+4d)⇒a+4d = 18⇒2d + 4d = 18     [Using (1)]⇒6d = 18⇒d = 3 Now, from (1), we get a = 2×3 = 6 So, first term, a = 6

we can verify our answer by given a and d
we know that A6:A13 is 1:2
So A6=6+(6-1)×3
A6 = 21
now A13=6+(13-1)×3
A13=6+12×3
A13=6+36
A13=42

so 21:42 is In ratio 1:2

I hope I would get brainlist answer award

Answer 2

S9=162
T6:T13=1:2
S9= a+a+d+a+2d+a+3d+a+4d+a+5d+a+6d+a+7d+a+8d=162
9a+36d=162-->1.
a+5d:a+12d=1:2
a+5d/a+12d=1/2
2a+10d=a+12d
a=2
Put, a=2d in eq. 1.
9(2d)+36d=162
18d+36d=162
54d=162
d=3
a=2d =>a=2(3) =>a=6
Therefore, a=6