Total surface area of a narrow hemisphere of
external and internal radii R and ris given by​

1 thought on “Total surface area of a narrow hemisphere of<br />external and internal radii R and ris given by​”

1. Step-by-step explanation:

   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

   Reply