Question

A student takes Copper, Aluminium, Iron and
Zinc metals separately in four test tubes
labelled as I, II, III and IV respectively, He adds
10 ml of freshly prepared Ferrous Sulphate
solution to each test tube andobserves the
colour of the metal residue in each case. He
would observe a brown residue in the test
tubes. Identify the test tubes with brown
residue.
(IV)
POBE
goed
PREVIOUS
SAVE NEXT​

Answer 1

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