The 16^{th} term of an A.P. is five times its third term. If its 10^{th} term is 41, then find the sum of its first fifteen terms.

#### Solution

Given that 16th term of an A.P. is five times its third term.

We know that

Thus,

`t_16=a+(16-1)d`

`t_3=a+(3-1)d`

Since `t_16= 5t_3 `, we have,

a+(16-1)d=5[a+(3-1)d]

a+15d=5[a+2d]

a+15d=5a+10d

5d=4a

4a-5d=0.......(1)

Also given that ` t_10= 41`

`t_10=a+(10-1)d`

41=a+9d

a+9d=41......(2)

Multiplying equation (2) by 4, we have,

4a + 36d = 164...(3)

Subtracting equation (1) from equation (3), we have,

[36-(-5)]d=164

41d=164

d=164/41

d=4

Substituting d = 4 in equation (1) 4a - 5d = 0,we have,

4a - 5 x 4=0

4a- 20= 0

4a=20

a=5

We need to find `S_15`

We know that

`S_n=n/2[2a+(n-1)d]`

`S_15=15/2[2xx5+(15-1)xx4][a=5,n=15,d=4]`

`S_15=15/2[10+14xx4]`

`S_15=15/2xx66`

`S_15=495`