Question

The sum of 4th and 8th terms of a.p.is 24 and the sum of 6th and 10th terms of same a.p.is 34.find the first three terms of a.p.

Answer 1

Given :

• The sum of 4th and 8th terms of a.p.is 24.
• The sum of 6th and 10th terms of same a.p.is 34.

To Find :

• First three term of AP.

Solution :

Let the first term of the AP be a.

Let the common difference be d.

Given that, sum of 4th term i.e $\sf{t_4}$ and 8th term, $\sf{t_8}$ is 24.

Fourth term of AP :

$\longrightarrow$ $\sf{t_4\:=\:a+(4-1)d}$

$\longrightarrow$ $\sf{t_4\:=\:a+(3)d}$

$\longrightarrow$ $\sf{t_4\:=\:a+\:3d}$

Eight term of AP :

$\longrightarrow$ $\sf{t_8\:=\:a+(8-1)d}$

$\longrightarrow$ $\sf{t_8\:=\:a+(7)d}$

$\longrightarrow$ $\sf{t_8\:=\:a+\:7d}$

Sum of 4th and 8th term is 24.

$\longrightarrow$ $\sf{t_4\:+\:t_8\:=\:24}$

Block in the value,

$\longrightarrow$ $\sf{a+3d+a+7d=24}$

$\longrightarrow$ $\sf{2a\:+\:10d\:=\:24\:\:(1)}$

Now, sum of 6th term and 10th term is 34.

6th term of AP :

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

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

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

10th term of AP :

$\longrightarrow$ $\sf{t_{10}\:=\:a+(10-1)d}$

$\longrightarrow$ $\sf{t_{10}\:=\:a+(9)d}$

$\longrightarrow$ $\sf{t_{10}\:=\:a+9d}$

Sum of $\sf{t_6\:\:and\:\:t_{10}\:=\:34}$

$\longrightarrow$ $\sf{t_6\:+\:t_{10}=34}$

$\longrightarrow$ $\sf{a+5d+a+9d=34}$

$\longrightarrow$ $\sf{2a+14d=34\:\:(2)}$

Now, subtract equation (1) from (2),

$\longrightarrow$ $\sf{2a+10d-(2a+14d)=24-34}$

$\longrightarrow$ $\sf{2a+10d-2a-14d=-10}$

$\longrightarrow$ $\sf{-4d=-10}$

$\longrightarrow$ $\sf{d=\dfrac{-10}{-4}}$

$\longrightarrow$ $\sf{d=\dfrac{10}{4}}$

Substitute, d = 10/4 in (2),

$\longrightarrow$ $\sf{2a+14d=34}$

$\longrightarrow$ $\sf{2a+14\:\times\:\dfrac{10}{4}=34}$

$\longrightarrow$ $\sf{2a+\dfrac{140}{4}=34}$

$\longrightarrow$ $\sf{2a+35=34}$

$\longrightarrow$ $\sf{2a=34-35}$

$\longrightarrow$ $\sf{2a=-1}$

$\longrightarrow$ $\sf{a=\dfrac{-1}{2}}$

$\longrightarrow$ $\sf{a=-0.5}$

We have the first term, a and the common difference, d of the AP.

•°• First three terms of the AP :

$\large{\boxed{\sf{First\:Term\:=\:a\:=\:-0.5}}}$

$\large{\boxed{\sf{Second\:term\:=\:t_2\:=\:a\:+\:d\:=\:-0.5+\dfrac{10}{4}\:=-0.5+2.5=\:2}}}$

$\large{\boxed{\sf{Third\:term\:=\:\:t_2\:+\:d\:=\:2+\dfrac{10}{4}\:=2+2.5=\:4.5}}}$

Answer 2

$\huge \mathtt{ \fbox{Solution :)}}$

Given ,

$\star$ The sum of 4th and 8th term of AP is 24

It can be written as ,

a + 3d + a + 7d = 24

2a + 10d = 24 ----- (1)

$\star$ The sum of 6th and 10th terms of AP is 34

It can be written as ,

a + 5d + a + 9d = 34

2a + 14d = 34 ----- (2)

Now , subtract eq (1) from eq (2) , we get

2a + 14d - (2a + 10d) = 34 - 24

4d = 10

d = 10/4

Put the value of d = 10/4 in eq (1) , we get

2a + 10(10/4) = 24

2a + 100/4 = 24

(8a + 100) = 96

8a = 96 - 100

a = - 4/8

a = - 0.5

Hence , the first three terms of AP are

• First term (a) = - 0.5
• Second term (a + d) = 2
• Third term (a + 2d) = 4.5

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