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

Answer:

The sum of first 50 terms is 3425.

Step-by-step 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

$S_n=\frac{n}{2}[2a+(n-1)d]\\\\5=\frac{5}{2}[2a+4d]\\\\1=a+2d..........(ii)$

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

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

Thus $a=1-2\times 3=-5$

Thus the sum of first 50 terms euals

$S_{50}=\frac{50}{2}[2\times -5+(50-1)3]\\\\S_{50}=3425.$

Answer 2

Answer:

$S_{50}=3425$

Step-by-step 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:

$S_n=\frac{n}{2}[2a_1+(n-1)d]$

$5=\frac{5}{2}[2a_1+(5-1)d]$

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

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

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

$40-15d=a_1$

Substitute this value in equation (2).

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

$5=200-75d+10d$

$5=200-65d$

$200-65d=5$

$200-200-65d=5-200$

$-65d=-195$

$\frac{-65d}{-65}=\frac{-195}{-65}$

$d=3$

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:

$S_{50}=\frac{n}{2}[2a_1+(n-1)d]$

$S_{50}=\frac{50}{2}[2(-5)+(50-1)3]$

$S_{50}=25[-10+(49)3]$

$S_{50}=25[-10+147]$

$S_{50}=25[137]$

$S_{50}=3425$

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