Question

Find an Ap whose fourth term is 9 and sum of sixth And 15th term is 40

Answer 1

Answer:

⇒ $\frac{51}{13} ,\frac{73}{13} , \frac{95}{13},..$

Step-by-step explanation:

Given :

Find an AP,

whose fourth term is 9

and

sum of sixth And 15th term is 40.

Solution :

Let the 1st term be a ,

&

Common difference be d,.

We know that,

$a_n=a+(n-1)d$

(given) 4th term of an AP is 9

⇒ $a_4 = 9$

⇒ $a_4 = a + (4 - 1)d = 9$

⇒ $a+3d = 9$ ..(i)

(given) Sum of sixth And 15th term is 40.

⇒ $a_6 + a_15 = 40$

⇒ $a+(6-1)d+a+(15-1)d = 40$

⇒ $a+5d+a+14d = 40$

⇒ $2a+19d = 40$ ...(ii)

__

By subtracting 2 × (i) from (ii)

⇒ $(2a+19d) - 2\times(a+3d) = 40 - 2 \times(9)$

⇒ $(2a+19d) - (2a+6d) = 40 - 18$

⇒ $2a+19d - 2a-6d = 22$

⇒ $13d = 22$

⇒ $d=\frac{22}{13}$

By substituting value of d in (i),

We get,

⇒ $a+3d= 9$

⇒ $a+3\times\frac{22}{13} = 9$

⇒ $a+\frac{66}{13} = 9$

⇒ $a = 9 - \frac{66}{13}$

⇒ $a =\frac{117-66}{13}$

⇒ $a =\frac{51}{13}$

Then, the AP is,

⇒ $a_1, a_2,a_3,..$

⇒ $a,a+d,a+2d,..$

⇒ $\frac{51}{13} ,\frac{51}{13} + \frac{22}{13} , \frac{51}{13} +2\times \frac{22}{13},..$

⇒ $\frac{51}{13} ,\frac{51+22}{13} , \frac{51}{13} + \frac{44}{13},..$

⇒ $\frac{51}{13} ,\frac{73}{13} , \frac{51+44}{13},..$

⇒ $\frac{51}{13} ,\frac{73}{13} , \frac{51+44}{13},..$

⇒ $\frac{51}{13} ,\frac{73}{13} , \frac{95}{13},..$