Question

did any pls solve it this question is related to AP....​

Answer 1

Answer:

hey here is your answer

so pls mark it as brainliest

so here we go

Step-by-step explanation:

now let the five terms of an A.P be a,a+d,a+2d,a+3d,a+4d respectively

so according to first condition

a+a+d+a+2d+a+3d+a+4d=55

ie 5a+10d=55

so a+2d=11

hence a=11-2d (1)

now according to second condition

a+3d=5+a+a+d

so a+3d=5+2a+d (2)

substitute value of a from (1) in (2)

we get

11-2d+3d=5+2(11-2d)+d

so 11+d=5+22-4d+d

11+d=27-3d

so 4d=16

thus d=4

substitute value of d in (1)

we get

a=3

so thus

a=3

a+d=(3+4)=7

a+2d=3+(2×4)

=3+8

=11

a+3d=(3+(3×4))

=3+12

=15

a+4d=(3+(4×4))

=3+16

=19

hence the required 5 terms of above AP are 3,7,11,15,19,and so on