Question

The ninth term of an AP is equal to seven times the

second term and twelfth term exceeds five times the

third term by 2. Find the first term and the common

difference.​

Answer 1

Answer:

• First term = 1
• Common difference = 6

Given statements about the terms of an AP:

• 9th term = 7 × 2nd Term
• 9th term = 7 × 2nd Term12th term = 5 × 3rd term + 2

We have to find the following:

1. First term, a
2. Common difference, d

The standard form of an AP is:

a , a + d, a + 2d , a + 3d, ... , a + (n - 1)d

Where,

• a = first term of AP
• d = common difference of AP

So, According to the formula,

• aₙ = a + (n - 1)d

We have 9th term and 2nd term as a + 8d and a + d respectively. So According to the statement given,

⇒ 9th term = 7 × 2nd term

⇒ a + 8d = 7 (a + d)

⇒ a + 8d = 7a + 7d

⇒ 7a - a + 7d - 8d = 0

⇒ 6a - d = 0 ...(i)

Similarly, According to the second statement, we have

⇒ 12th term = ( 5 × 3rd term ) + 2

⇒ a + 11d = { 5(a + 2d) } + 2

⇒ a + 11d = 5a + 10d + 2

⇒ 5a - a + 10d - 11d = -2

⇒ 4a - d = -2 ...(ii)

Subtract eq.(ii) from eq.(i), we get

⇒ 6a - d - (4a - d) = 0 - (-2)

⇒ 6a - d - 4a + d = 2

⇒ 6a - 4a = 2

⇒ 2a = 2

⇒ a = 1

We found the first term to be 1, Hence substitute the value of a in eq.(i), we get

⇒ 6a - d = 0

⇒ 6(1) - d = 0

⇒ 6 - d = 0

⇒ d = 6

Answer 2

Answer:

Given :-

• 9th term of AP = 7 times 2nd term
• If twelfth term exceeds five times the third term by 2.

To Find :-

First and common Difference

Solution :-

Let the terms be

a

a + d

a + 2d

a + 3d...

Nth term = a + (n - 1)d

$\sf \: a \: + 8d =7 \times(a + d)$

$\sf \: a + 8d = 7a + 7d$

$\sf 7a - a = 7d - 8d$

$\sf \: 6a - d = 0$

For second condition

$\sf \: a + 11d = 5 \times (a + 2d) + 2$

$\sf \: a + 11d = 5a + 10d + 2$

$\sf \: 5a - a = 11d - 10d - 2$

$\sf \: 4a + d = 2$

By substituting the values

$\sf \: 6a -d - (4a + d) = 0 - (-2)$

$\sf \: 6a - 4a \: ( - d + d) = 2$

Cancelling d

$\sf \: 6a - 4a = 2$

$\sf \: 2a = 2$

$\sf \: a \: = \dfrac{2}{2}$

$\sf \: a \: = 1$

First term = 1

$\sf \: 6a - d = 0$

$\sf \: 6 \times 1 - d = 0$

$\sf \: 6 - d = 0$

$\sf \: - d = - 6$

$\sf \: d \: = 6$