Question

9th term of AP is equal to 7 times the second term and 12th term exceeds 5 times the 3rd term by 2. Find AP​

Answer 1

Step-by-step explanation:

Let ' a ' be the first term and ' d ' be the common difference.

Case 1:-

9th term = 7( 2nd term)

➝ $\rm a_{9} = 7( a_{2})$

➝ a + 8d = 7( a + d)

➝ a + 8d = 7a + 7d

➝ 7a - a = 8d - 7d

➝ 6a = d.....(i)

Case 2:-

12th term = 5(3rd term) + 2

➝ $\rm a_{12} = 5( a_{3})+ 2$

➝ a + 11d = 5(a + 2d) + 2

➝ a + 11d = 5a + 10d + 2

➝ 11d - 10d = 5a - a + 2

➝ d = 4a + 2

➝ 6a = 4a + 2 $\: \: \: \: \:$ [ d = 6a]

➝ 2a = 2

➝ a = 1

➝ d = 6a = 6 × 1 = 6

We have:-

• a = 1

• d = 6

First term = a = 1

Second term = a + d = 1 + 6 = 7

Third term = a + 2d = 1 + 2(6) = 13

Fourth term = a + 3d = 1 + 3(6) = 19

Therefore:-

The AP = 1 , 7 , 13 , 19.......

Answer 2

Step-by-step explanation:

Let ' a ' be the first term and ' d ' be the common difference.

Case 1:-

9th term = 7( 2nd term)

➝

$\rm a_{9} = 7( a_{2})a$

➝ a + 8d = 7( a + d)

➝ a + 8d = 7a + 7d

➝ 7a - a = 8d - 7d

➝ 6a = d.....(i)

Case 2:-

12th term = 5(3rd term) + 2

➝

$\rm a_{12} = 5( a_{3})+ 2a$

➝ a + 11d = 5(a + 2d) + 2

➝ a + 11d = 5a + 10d + 2

➝ 11d - 10d = 5a - a + 2

➝ d = 4a + 2

➝

$6a = 4a + 2 \: \: \: \: \: [ d = 6a]$

➝ 2a = 2

➝ a = 1

➝ d = 6a = 6 × 1 = 6

We have:-

a = 1

d = 6

First term = a = 1

Second term = a + d = 1 + 6 = 7

Third term = a + 2d = 1 + 2(6) = 13

Fourth term = a + 3d = 1 + 3(6) = 19

Therefore:-

The AP = 1 , 7 , 13 , 19.......