Question

if c+2,4c-6 and 3c-2 are three consecutive terms of an AP, then find the value of c.​

Answer 1

EXPLANATION.

c + 2, 4c - 6 and 3c - 2 are three consecutive terms,

As we know that,

METHOD = 1.

Common difference of an A.P. = d = b - a = c - b.

⇒ (4c - 6) - (c + 2) = (3c - 2) - (4c - 6).

⇒ 4c - 6 - c - 2 = 3c - 2 - 4c + 6.

⇒ 3c - 8 = - c + 4.

⇒ 3c + c = 4 + 8.

⇒ 4c = 12.

⇒ c = 12/4.

⇒ c = 3.

METHOD = 2.

As we know that,

Conditions of an A.P.

⇒ 2b = a + c.

⇒ 2(4c - 6) = c + 2 + (3c - 2).

⇒ 8c - 12 = c + 2 + 3c - 2.

⇒ 8c - 12 = 4c.

⇒ 8c - 4c = 12.

⇒ 4c = 12.

⇒ c = 12/4.

⇒ c = 3.

                                                                                                                       

MORE INFORMATION.

General term of an A.P.

General term (nth term) of an A.P is given by,

Tₙ = a + (n - 1)d.

Sum of n terms of an A.P.

Sₙ = n/2[2a + (n - 1)d]  Or  Sₙ = n/2[a + Tₙ].

(1) = If sum of n terms Sₙ is given then general term Tₙ = Sₙ - Sₙ₋₁ where Sₙ₋₁ is sum of (n - 1) terms of A.P.

Answer 2

Answer:

$\implies\boxed{\bf{c = 3}}$

Step-by-step explanation:

Question:

If c+2,4c-6 and 3c-2 are three consecutive terms of an AP, then find the value of c.​

Concept involved:

Here, the concept of arithmetic progression is involved. We need to find the value of c for the given question.

Let's do this!

A.P = c+2,4c-6, 3c-2

Here,

• a = c+2
• $\bf{a_2}$ = 4c-6
• $\bf{a_3}$ = 3c-2

Common difference (or) d = $\bf{a_2-a = a_3 - a_2}$

Apply the values:

d = $\bf{4c-6-(c+2) = 3c-2 - (4c-6)}$

$\implies$  $\bf{4c-6-c-2 = 3c-2 -4c+6}$

$\implies$  $\bf{3c-8 = -c+4}$

$\implies$  $\bf{4c=12}$

$\implies\boxed{\bf{c = 3}}$