Question

The 16th term of an ap is 5 times its 3rd term. If its 10th term is 41 then find the sum of its first 15 terms

Answer 1

Hello !

16th term = a + 15d
3rd term = a + 2d

given

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

4a - 5d = 0

4a = 5d

a = 5d/4

Also,

10th term = a + 9d = 41


a = 5d/4

so,

5d/4 + 9d = 41

5d/4 + 36d/4 = 41

5d + 36d/4 = 41

41d/4 = 41

41d = 164

d = 164/41

d = 4



a + 9d = 41

a + 36 =41
a = 5

---------------------------------------------

n = 15
a = 5
d = 4

Sn = n/2(2a + (n-1)d)
= 15/2 (10 + 14 * 4)
= 15/2*66
= 495

Answer 2

Answer:

Step-by-step explanation:

a16=5×a3

a10= 41

a16 = 5(a+2d).

a+9d=41....1

= 4a+36=164......2

a+15d=5a+10d

4a-5d....3

3-2

41d=164.

D=4

S15 =15/2(20+14×4)

15/2(66)

495