Question

the 16th term of an AP is 5times it's 3rd term, if it's 10th term is 41,find the sum of its first 15 terms ​

Answer 1

Answer:

16th term of the AP = a+15d

3rd term of the AP = a+2d

so, a+15d = 5(a+2d) = 5a+10d

that is, 15d - 10d = 5a - a

that is, 5d = 4a

that is, a=5d/4

10th term of the AP = 41 =a+9d

substituting a as 5d/4 ,we get

5d/4+9d = 41

that is, 41d/4 = 41

that is, d = 41×4/41 = 4

now,a =5d/4 = (5×4/4)

=5

sum of first 15 terms

= n/2(2a+(n-1)d)

substituting the values ,we get

15/2(2×5 + (15-1)×4)

=15/2 (10 + 56) = 15/2 × 66

= 15 × 33 = 495