Question

The sum of 4th and 8th AP term is 24 and 6th and 10th term is 44 write the first three term of AP

Answer 1

EXPLANATION.

• GIVEN

Sum of 4th and 8th term of an Ap = 24

sum of 6th and 10th term = 44

Find the first three term in Ap

According to the question,

Nth term of an Ap

=> An = a + ( n - 1 ) d

Case = 1

4th and 8th term = 24

=> a + 3d + a + 7d = 24

=> 2a + 10d = 24

=> a + 5d = 12 ......(1)

Case = 2

6th and 10th term = 44

=> a + 5d + a + 9d = 44

=> 2a + 14d = 44

=> a + 7d = 22 ..... (2)

From equation (1) and (2) we get,

=> -2d = -10

=> d = 5

put the value of d = 5 in equation (1)

we get,

=> a + 25 = 12

=> a = -13

Therefore,

First term = a = -13

common difference = d = 5

First three term

First term = a = -13

second term = a + d = -13 + 5 = -8

third term = a + 2d = -13 + 2(5) = -3

First three term = -13 , -8 , -3

Answer 2

Step-by-step explanation:

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

• Sum of 4th and 8th term of an AP is 24

• And sum of 6th and 10th terms is 44

$\bf {\underline{\underline\blue{To\:Find:-}}}$

• The first three terms of the AP

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

As we know that:-

The nth term of an AP is given by the formula

$\boxed{ \rm a_{n} = a + (n - 1)d}$

Here:-

• $\rm a_{n} = nth \: term$

• $\rm a = 1st \: term$

• $\rm n \: = number \: of \: terms$

• $\rm d \: = common \: difference$

Sum of 4th term and 8th term = 24

$\rm \longrightarrow a_{4} + a_{8} = 24$

$\rm \longrightarrow \{ a + (4 - 1)d \}+ \{ a + (8 - 1)d \} = 24$

$\rm \longrightarrow a +3d + a + 7d= 24$

$\rm \longrightarrow 2a + 10d = 24$

$\rm \longrightarrow a + 5d = 12.......(i)$

Sum of 6th term and 8th term = 44

$\rm \longrightarrow a_{6} + a_{10} = 44$

$\rm \longrightarrow \{ a + (6 - 1)d \}+ \{ a + (10 - 1)d \} = 44$

$\rm \longrightarrow a +5d + a + 9d= 44$

$\rm \longrightarrow 2a + 14d = 44$

$\rm \longrightarrow a + 7d = 22.......(ii)$

Adding equations (i) and (ii)

• a + 5d = 12
• a + 7d = 22

Eliminating (a) we get

$\rm \longrightarrow 7d - 5d = 22 - 12$

$\rm \longrightarrow 2d = 10$

$\rm \longrightarrow d = \dfrac{10}{2}$

$\rm \longrightarrow d = 5$

Substituting d = 5 in (i)

$\rm \longrightarrow a + 5d = 12$

$\rm \longrightarrow a + 5(5) = 12$

$\rm \longrightarrow a + 25 = 12$

$\rm \longrightarrow a = 12 - 25$

$\rm \longrightarrow a = - 13$

Therefore:-

• The first term = a = -13

• Second term = a + d = -13 +5 = -8

• Third term = a + 2d = -13 + 2(5) = -13 + 10 = - 3

Hence;

• First term = -13

First term = -13• Second term = -8

First term = -13• Second term = -8• Third term = -3