Question

The sum of the 4th and 8th term of an ap is 24 and the sum of the 6th and 10th term is 44. find the first three term of AP​

Answer 1

Answer:

a+3d+a+7d=24. a+5d+a+9d=44. *a=-13

2a+10d=24. 2a+14d=44. a+d=-8

2(a+5d)=24. 2(a+7d)=44. a+2d=-3

a+5d=12.-(1) a+7d=22-(2)

Step-by-step explanation:

solve (1) and (2) 2d=10. d=5 sub d in (1) a=-13

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Given:

Sum of 4th and 8th term = 24

Sum of 6th and 10th term = 44

Assume,

First term be p and Common difference be n

So,

p₄ + p₈ = 24

p + 3n + p + 7n = 24

2p + 10n = 24 ......(1)

Now,

p₆ + p10 = 44

p + 5n + p + 9n = 44

2p + 14n = 44 ......(2)

Solving it further,

2p + 10n = 24

2p = 24 - 10n

$\rm \: \: \: \: \: \: \: \: \:p=\dfrac{24-10n}{2}$

Put the value of 'p' in (2),

2p + 14n = 44

$\rm \: \: \: \: \: \: \: \: \:2\times\dfrac{24-10n}{2}+14n=44$

$\rm \: \: \: \: \: \: \: \: \:\dfrac{48-20n}{2}+14n=44$

$\rm \: \: \: \: \: \: \: \: \:2\times\dfrac{24-20n+28n}{2}=44$

48 - 20n + 28n = 88

8n = 40

$\rm \: \: \: \: \: \: \: \: \:n=\dfrac{\cancel{40}}{\cancel{8}}$

n = 5

Put the value of 'n' in (1),

2p + 10n = 24

2p + 50 = 24

2p = -26

p = - 13

First three terms,

p₁ = -13

p₂ = p + n = -13 + 5 = -8

p₃ = p + 2n = -13 + 10 = -3

Hence,

First three terms of AP are -13, -8, -3