Question

Q(4) The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th

and 10th terms is 44. Find the first two terms of the A.P.​

Answer 1

4term is =a+3d

8term is =a+7d

Sum of 4term and 8term is

a+3d+a+7d=2a+10d

2a+10d=24

a+5d=12

a=12-5d

6term is a+5d

10term is a+9d

Sum of 6term and 10term is

a+5d+a+10d=2a+15d

2a+15d=44

2(12-5d)+15d=44

24-10d+15d=44

24+5d=44

5d=20

d=4

a=12-5d

a=12-5(4)=12-20=-8

First term is - 8,common difference is 4

A.Pis

-8,-8+4,-8+4+4,-8+4+4+4......

-8,-4,0,+4......

Answer 2

SOLUTION :-

We know that,

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

$\bullet\sf \ a_4+a_8 = 24$

  $\longrightarrow\sf [ \ a+(4-1)d \ ] + [ \ a+(8-1)d \ ] = 24 \\\\\longrightarrow (a+3d)+(a+7d) = 24 \\\\\longrightarrow 2a+10d = 24 \\\\\longrightarrow a+5d = 12 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...................(1)$

$\bullet\sf \ a_6+a_{10}= 44$

  $\longrightarrow\sf [ \ a+(6-1)d \ ] + [ \ a+(10-1)d \ ] = 44 \\\\\longrightarrow (a+5d)+(a+9d)= 44 \\\\\longrightarrow 2a+14d = 44 \\\\\longrightarrow a+7d = 22 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...................(2)$

Eq (2) - Eq (1),

    $\longrightarrow\sf 2d = 10 \\\\\longrightarrow\bf d= 5$

Substitute the value of d in eq (1),

   $\longrightarrow\sf a+5\times 5 = 12 \\\\\longrightarrow a+25 = 12 \\\\ \longrightarrow\bf a = -13$

First term = -13

Second term = -8