Question

solve this in paper and send please​

Answer 1

$\huge \boxed{hey \: there}$

Given:

a4+a8=24

Now,

➡a+3d+a+7d=24

➡2a+10d=24

➡a+5d=12-----------(1)

Also,

➡a6+a10=44

➡a+5d+a+9d=44

➡2a+14d=44

➡a+7d=22------------(2)

Subtracting equation (2) from (1),

↪ -2d= -10

↪ d=5

Substituting this value of d in eq(1),

↪a+5(5)=12

↪a+25=12

↪a=12-25

↪a= -13

a=-13 and d=5

Hence, the first three terms of the AP will be:

↪a=-13

↪a+d=-13+5=-8

↪a+2d=-13+2(5)=-13+10=-3

➡-13,-8,-3..

Answer 2

Let the first term be 'a' and common difference be 'd' .

Given ,
4th term + 8th term = 24
=> a + 3d + a +7d = 24
=> 2a + 10d = 24
=>2 (a+5d)= 2×12
=> a + 5d = 12 ------------> i

and ,
6th term + 10th term = 44
=> a+ 5d + a + 9d = 44
=> 2a + 14d = 44
=> a + 7d = 22---------> ii

Now ,
ii- i
=> a+7d -(a + 5d) = 22-12
=> a + 7d - a - 5d = 10
=> 2d = 10
=> d = 5

again ,
using d = 5 in (i) we have ,
=> a + 5×5 = 12
=> a = 12 - 25
=> a = -13

So , terms of AP are
-13 ,-13+5 , -13 + 2×5 , -13 + 3×5 , .................-13 + (n -1) 5

= -13 ,-8,-3,2 ,..............-13+(n-1)5