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

second term and twelfth term exceeds five times the

th

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.​

Savannah · Answer

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:

First term, a

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

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Emma · Answer