Question

The sum of 3rd and 6th term of an A.P. is 20 and the sum of 6th and 5th term is 40 find the first 5 terms of an A.P. Please answer quickly!!!!!!​

Answer 1

Answer:

• First five terms of AP are -25, -15, -5, 5 and 15.

Step-by-step explanation:

Given:

• Sum of 3rd and 6th term = 20
• Sum of 6th and 5th term = 40

To Find:

• First 5 terms of AP.

Now, according to question.

=> a₃ + a₆ = 20

=> a + 2d + a + 5d = 20

=> 2a + 7d = 20 .......(1)

And,

=> a₆ + a₅ = 40

=> a + 5d + a + 4d = 40

=> 2a + 9d = 40 .......(2)

Now, we will solve these equations by substitution method.

=> 2a + 7d = 20

=> 2a = 20 - 7d

=> a = (20 - 7d)/2

Now, put the value of 'a' in equation (2).

=> 2a + 9d = 40

=> 2[(20 - 7d)/2] + 9d = 40

=> [(40 - 14d)/2] + 9d = 40

=> (40 - 14d + 18d)/2 = 40

=> 40 + 4d = 80

=> 4d = 40

=> d = 40/4

=> d = 10

Now, put the value of 'd' in equation (1).

=> 2a + 7d = 20

=> 2a + 7(10) = 20

=> 2a + 70 = 20

=> 2a = 20 - 70

=> 2a = -50

=> a = -50/2

=> a = -25.

Hence, First five terms are:

• a₁ = - 25
• a₂ = - 25 + 10 = - 15
• a₃ = - 25 + 20 = -5
• a₄ = - 25 + 30 = 5
• a₅ = - 25 + 40 = 15

Answer 2

Given :

• The sum of 3rd and 6th term of an A.P. is 20
• The sum of 6th and 5th term is 40.

To Find :

• First 5 terms of the AP.

Solution :

Let the first term be a.

Let the common difference be d.

Case 1 :

The sum of 3rd term i.e $\sf{t_3}$ and 6th term i.e $\sf{t_6}$ is 20.

Third term of AP :

$\longrightarrow$ $\sf{t_3\:=\:a+(3-1)d}$

$\longrightarrow$ $\sf{t_3=a+(2)d}$

$\longrightarrow$ $\sf{t_3=a+2d\:\:(1)}$

Sixth term of AP :

$\longrightarrow$ $\sf{t_6\:=\:a+(6-1)d}$

$\longrightarrow$ $\sf{t_6=a+(5)d}$

$\longrightarrow$ $\sf{t_6=a+5d\:\:(2)}$

The sum of 3rd and 6th term is 20.

$\longrightarrow$ $\sf{t_3\:+\:t_6=20}$

$\longrightarrow$ $\sf{a+2d+a+5d=20}$

$\longrightarrow$ $\sf{2a+7d=20\:\:\:(3)}$

Case 2 :

The sum of 6th term and 5th term is 40.

Sixth term of AP :

$\longrightarrow$ $\sf{t_6=a+5d}$

$\sf{\big[From\:equation\:(2)\big]}$

Fifth term of AP :

$\longrightarrow$ $\sf{t_5\:=\:a+(5-1)d}$

$\longrightarrow$ $\sf{t_5=a+(4)d}$

$\longrightarrow$ $\sf{t_5=a+4d\:\:\:(4)}$

Sum of 6th and 5th term is 40.

$\longrightarrow$ $\sf{t_6\:+\:t_5=40}$

$\longrightarrow$ $\sf{a+5d+a+4d=40}$

$\longrightarrow$ $\sf{2a+9d=40\:\:\:(5)}$

Now, subtract equation (3) from (5),

$\longrightarrow$ $\sf{2a+7d-(2a+9d)=20-40}$

$\longrightarrow$ $\sf{2a+7d-2a-9d=-20}$

$\longrightarrow$ $\sf{7d-9d=-20}$

$\longrightarrow$ $\sf{-2d=-20}$

$\longrightarrow$ $\sf{d=\dfrac{-20}{-2}}$

$\longrightarrow$ $\sf{d=\dfrac{20}{2}}$

$\longrightarrow$ $\sf{d=10}$

Substitute, d = 10 in equation (5),

$\longrightarrow$ $\sf{2a+9d=40}$

$\longrightarrow$ $\sf{2a+9(10)=40}$

$\longrightarrow$ $\sf{2a+90=40}$

$\longrightarrow$ $\sf{2a=40-90}$

$\longrightarrow$ $\sf{2a=-50}$

$\longrightarrow$ $\sf{a=\dfrac{-50}{2}}$

$\longrightarrow$ $\sf{a=-25}$

FIRST 5 TERMS :

$\large{\boxed{\sf{First\:Term\:=\:a\:=\:-\:25}}}$

$\large{\boxed{\sf{Second\:term\:=\:t_2\:=\:a+\:d=\:-\:25\:+\:10\:=\:-15}}}$

$\large{\boxed{\sf{Third\:term\:=\:t_3\:=\:t_2\:+\:d\:=\:-15\:+\:10\:=\:-5}}}$

$\large{\boxed{\sf{Fourth\:Term\:=\:t_4\:=\:t_3\:+d\:=\:-5+10\:=\:5}}}$

$\large{\boxed{\sf{Fifth\:term\:=\:t_5\:=\:t_4\:+\:d\:=\:5+10\:=\:15}}}$