Question

The 16th term of an A.P. is 40 and the sum of the first 5 terms is 5. Find the sum of the first 50 terms.

Answer 1

`S_(50)=3425`

Explanation:

We have been that the 16th term of an A.P. is 40 and the sum of the first 5 terms is 5.

We will use arithmetic sequence formula and arithmetic sequence sum formula to solve our given problem.

Sequence formula:

`a_n=a_1+(n-1)d`, where,

n = Number of terms in a sequence,

d = Common difference.

`40=a_1+(16-1)d`

`40=a_1+15d...(1)`

Sum formula:

`S_n=(n)/(2)[2a_1+(n-1)d]`

`5=(5)/(2)[2a_1+(5-1)d]`

`5=2.5[2a_1+4d]`

`5=5a_1+10d...(2)`

Now, we have two unknown and two equations. From equation (1), we will get:

`40-15d=a_1`

Substitute this value in equation (2).

`5=5(40-15d)+10d`

`5=200-75d+10d`

`5=200-65d`

`200-65d=5`

`200-200-65d=5-200`

`-65d=-195`

`(-65d)/(-65)=(-195)/(-65)`

`d=3`

Substitute
`d=13` in equation (1):

`40=a_1+15(3)`

`40=a_1+45`

`40-45=a_1+45-45`

`-5=a_1`

Use sum formula to find sum of first 50 terms:

`S_(50)=(n)/(2)[2a_1+(n-1)d]`

`S_(50)=(50)/(2)[2(-5)+(50-1)3]`

`S_(50)=25[-10+(49)3]`

`S_(50)=25[-10+147]`

`S_(50)=25[137]`

`S_(50)=3425`

Therefore, the sum of first 50 terms of the given sequence would be 3425.

Answer 2

The sum of first 50 terms is 3425.

Explanation:

Let the first tem of A.P be 'a' and it's common difference be 'd'

Thus the 16th term of the A.P is given by

`T_n=a+(n-1)d\\\\40=a+15d..........(i)`

Now we know that the sum of first 'n' terms of the A.P is given by

`S_n=(n)/(2)[2a+(n-1)d]\\\\5=(5)/(2)[2a+4d]\\\\1=a+2d..........(ii)`

Solving equation 'i' and 'ii' simultaneously we get

`40=1-2d+15d\\\\13d=39\\\\\therefore d=3`

Thus
`a=1-2* 3=-5`

Thus the sum of first 50 terms euals

`S_(50)=(50)/(2)[2* -5+(50-1)3]\\\\S_(50)=3425.`