Question

The 7th term of an AP is 4 times its 2nd term and 12 th term is 2 more than 3 times of its 4th term find progression

Answer 1

Answer:

2,5,8,..

Step-by-step explanation:

Solve by following the process I have shown. Hope it helps you

Answer 2

$\sf\bold{\pink{QմєรƬเoɳ:-}}$

$\sf\small{\green{\:The \: 7th\: term\: of \: an\: AP\: is\: four\: times\: its\: 2nd\: term\: is\: two\: more\: than\: three\: times\: of\: its\: 4 th\: term. \: Find\: the\: progression.}}$

$\sf\bold{\pink{卂иรωεƦ:-}}$

The progression is :2,5,8,11,14,17

$\sf\bold{\pink{\: Step\: by\: Step\: Explanation:-}}$

$\sf\small{\blue{\:Given \: that:-}}$

The 7th term of an AP is four times its 2nd term.

12th term is two more than three times its 4 th term.

$\sf\small{\blue{\: To\: find:-}}$

The progression.

$\sf\small{\blue{\:Let \: us \: Assume:-}}$

First term of an AP = a

Common difference = d

$\sf\small{\blue{\:We\: know\: that:-}}$

⟹Tn = a +(n - 1)×d⟹Tn=a+(n−1)×d

$\sf\small{\orange{where,}}$

Tn = nth term

a = First term

n = Number of terms

d = Common difference

$\sf\small{\blue{\: According\: to\: question:-}}$

The seventh term of an AP is four times its 2nd term.

⟹T₇ = 4 T₂

⟹a + (7 - 1)d = 4{a+(2 -1)d}

⟹a + 6d = 4{a+d}

⟹4a - a = 6d - 4d

⟹3a = 2d_______(i)

12th term is 2 more than three times of its 4th term.

⟹T₁₂ = 3T₄ + 2

⟹a + (12 - 1) d =3{a + (4 - 1)d}+2

⟹a+11d = 3{a +3d}+2

⟹a + 11 d = 3a + 9d + 2

⟹11d - 9d = 3a - a + 2

⟹2d = 2a + 2⠀⠀⠀⠀ [From eqⁿ (i)2d = 3a]

⟹3a = 2a + 2

⟹3a - 2a = 2

⟹a = 2

⠀⠀⠀⠀First term= 2

In equation (i),

⟹2d = 3a

⟹2d = 3×2⠀⠀⠀⠀⠀ [ The value of a = 2]

⟹2d = 6

⟹d = 3

⠀⠀⠀⠀Common difference = 3

$\sf\small{\blue{\:Some \: of \:AP:-}}$

⟹T₁ = a = 2

⟹T₂ = a + d = 2 + 3 = 5

⟹T₃ = a + 2d = 2 + 6 = 8

⟹T₄ = a + 3d = 2 + 9 = 11

⟹T₅ = a + 4d = 2 + 12 = 14

⟹T₆ = a + 5d = 2 + 15 = 17

$\sf\bold{\green{\:The\: progression\: is \:2,5,8,11,14,17...}}$