Question

Sum of first 9 terms of an

a.p is 162. the ratio of 6th term to 13th term us 1:2. find the first and 15 th term of

a.p.

Answer 1

Hi Mate !!


Here's the Solution :-

Let first term be a
common difference be d


• Sum of 9 terms is 162

$sn \: = \frac{n}{2} (2a + (n - 1)d)$


$s9 = \frac{9}{2} (2a + 8d)$


$s9 = \frac{9 \times 2( a + 4d)}{2}$


$s9 = 9(a + 4d)$


$162 = 9(a + 4d)$


$18 = a + 4d \: \: \: .....(i)$

• Ratio of 6th term with 13th Term is 1:2

an = a + ( n - 1 )d

a6 = a + 5d

a13 = a + 12d


$\frac{a + 5d}{a + 12d} = \frac{1}{2}$


2a + 10d = a + 12d

2a - a + 10d - 12d = 0

a - 2d = 0 ........ ( ii )


Sub ( ii ) from ( i )

a + 4d - a + 2d = 18

6d = 18

d = 18/6

d = 3


Putting value of d in ( ii )

a - 2d = 0

a - 2 × 3 = 0

a - 6 = 0

a = 6


So, the first term ( a ) = 6

• 15 th term


a15 = a + 14d

a15 = 6 + 14 × 3

a15 = 6 + 42

a15 = 48