Question

find the sum of 25th term of the list of numbers whose nth term is an=5+2n​

Answer 1

Answer:

Given :t_n = 5 - 2nt

n

=5−2n

To Find: Find the sum of first 25 terms of an AP whose nth term is given by tn = 5- 2n

Solution:

t_n = 5 - 2nt

n

=5-2n

Put n =1

t_1 = 2 - 3(1)t

1

=2−3(1)

t_1 = -1t

1

=−1

put n =2

t_2 = 2 - 3(2)t

2

=2−3(2)

t_2= -4t

2

=−4

put n =3

t_3 = 2 - 3(3)t

3

=2−3(3)

t_3= -7t

3

=−7

So, A.P. become s: -1 , -4 , -7, ........

So, first term =a= -1

Common difference d = -4-(-1)=-7-(-4)= -3

Formula of sum of first n terms : \frac{n}{2}(2a+(n-1)d)

2

n

(2a+(n−1)d)

Put n =25

\frac{25}{2}(2(-1)+(25-1)(-3))

2

25

(2(−1)+(25−1)(−3))

\frac{25}{2}(-2-72)

2

25

(−2−72)

\frac{25}{2}(-74)

2

25

(−74)

-925−925

Hence the sum of first 25 terms of an AP is -925

Answer 2

Step-by-step explanation:

an=5+2n

for n=1

a1=5+2(1)

=5+2

=7

hence the first term is 7

for n=2

a2=5+2(2)

=5+4

=9

hence second term is 9

d=a2-a1

=9-7

=2

hence the common difference is 2

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

S25=25/2(2×7+2(25-1))

=25/2(14+2(24))

=25/2(14+48)

=25/2(62)

=25×31

=775

hence the sum of the first 25 terms is 775