Question

the sum of 4th and 8th term is 24the sum of the 6th and 10th term is 44 find the first three terms of the ap?​

Answer 1

SOLUTION :-

Let,

First term = a

Common difference = d

We know that,

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

$\bullet\sf \ 4^{th} \ term + 8 ^{th} \ term = 24$

  $\longrightarrow \sf a_4+a_8 = 24 \\\\\longrightarrow [ \ a+(4-1)d \ ] + [ \ a+ ( 8 -1 )d \ ] = 24 \\\\\longrightarrow (a+3d)+(a+7d) = 24 \\\\\longrightarrow 2a+10d = 24 \\\\\longrightarrow a+5d = 12 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(1)$

$\bullet \sf \ 6^{th} \ term + 10^{th} \ term = 44$

  $\longrightarrow \sf a_6+a_{10}= 44 \\\\\longrightarrow [ \ a+(6-1)d \ ] + [ \ a+(10-1)d \ ] = 44 \\\\\longrightarrow (a+5d)+(a+9d) = 44 \\\\\longrightarrow 2a+14d = 44 \\\\\longrightarrow a + 7d = 22 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ......................(2)$

Eq (2) - Eq (1),

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

Substitute the value of d in eq (1),

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

$\bullet \sf \ a_1= -13 \\\\\bullet \ a_2 = -13+5 = -8 \\\\\bullet \ a_3 = -8 + 5= -3$

First 3 terms of the AP are -13, -8 and -3.