Question

find the first four terms of A.P whose 7th term is 28 and 12th term is 43​

Answer 1

Answer:

a7=28

a7=a+6d

28=a+6d (eqn1)

a12=a+11d

43=a+11d(eqn2)

eqn1-2

-5d=-15

   d= -15/-5

     d= 3

substituting the value in eqn1

a+6d=28

a+6×3=28

a=28-18

a=10

a2=a+d

a2=10+3

a2=13

a3=a2+d

a3=13+3

a3=16

a4=a3+d

a4=16+3

a4=19

 a1=10

  a2=13

a3=16

a4=19

Answer 2

$\bf{\underline{Question:-}}$

find the first four terms of A.P whose 7th term is 28 and 12th term is 43.

$\bf{\underline{Given:-}}$

• 7th term of an A.P = 28
• 12th term of A.P = 43

$\bf{\underline{TO\:FIND:-}}$

• 1st four term = ?

$\bf{\underline{Solution:-}}$

Let,

• The 1st four term are = a , a1 + d, a2 + d and a3 + d

→ $\sf a_7 = a+(7-1)d = 28$

→$\sf a_7 = a + 6d = 28 ---eq(¡)$

→$\sf a_{12}= a+(12-1)d=43$

→$\sf a_{12} = a+11d = 43---eq(¡¡)$

★ Subtracting eq(¡¡) from eq(¡)

→$\sf a_{12} = a+11d = 43$

→$\sf{\underline{ a_7 = _-a + _-6d =_- 28}}$

→ 5d = 15

→ d = 15/5

→ d = 3

Substitute value of d in eq(¡)

→ $\sf a + 6d = 28$

→ $\sf a + 6 × 3 =28$

→ $\sf a = 28 - 6 × 3$

→ $\sf a = 10$

Now,

we have

• First term ( a ) = 10
• common difference ( d ) = 3

★ Finding 1st four term ★

Note ,

• Above we considered 1st four term

So,

• 1st term = a = 10
• 2nd term = a1 + d = 10 + 3 = 13
• 3rd term = a2 + d = 13 + 3 = 16
• 4th term = a3 + d = 16 + 3 = 19

$\bf{\underline{Hence:-}}$

• 1st four term are = 10 , 13 , 16 and 19

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