Question

The sum of the first 10 terms of an Ap is 145 and the sum of its fourth and ninth is 5 times the third term determine the first term and the constant difference

Answer 1

10/2 (2a+9d) = 145
a+3d + a+8d = 5(a+2d)

Answer 2

the answer is wrong we need correct answer for this question

Answer 3

S10=145
T4+T9=5×T3
therefore we can write this as
(a+3d)+(a+8d)=5(a+2d)
2a+11d=5a+10d
d=3a

to find the first term we can substitute our S10 & common difference into Sn=n/2[2a+(n-1)d] .

S10=10/2[2a+(10-1)3a]
145=5(2a+30a-3a)
145=5(29a)
145=145a {divide both sides by 145}
.•. a = 1

substitute a in d=3a to find the difference.

d=3(1)
d=3