Question

the fourth term of an ap is 10 if the 11th term is one more three times of the fourth term find the sum of its 25 terms

Answer 1

a+10d=a+3d+1

a+3d=10 _______(1)

then ,

a+10d=10×3+1

a+10d=31

-a-3d=-10

_____________

7d=21

d=3

put d =3 in eq 1

a+3d=10

a+3(3)=10

a+9=10

a=1

a25=a+24d

a25=1+24(3)

a25=73

Answer 2

Answer:

1037.5

Step-by-step explanation:

Formula of nth term in A.P. = $a_n=a+(n-1)d$

Substitute n = 4

So, $a_4=a+(4-1)d$

$a_4=a+3d$

we are given that the fourth term of an ap is 10 .

So, $a_4=a+3d =10$  --1

Substitute n = 11

$a_{11}=a+(11-1)d$

$a_{11}=a+10d$

Now we are given that  11th term is one more three times of the fourth term

So, $a_{11}=3a_4 +1$

So,  $a+10d=3(10) +1$

$a+10d=31$  --2

Subtract 1 form 2

$a+10d-a-4d=31-10$

$6d=21$

So, d = 21/6 =3.5

Substitute value of d in 1

$a+3(3.5) =10$

$a+10.5=10$

$a=-0.5$

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

Substitute n = 25

So, $S_{25}=\frac{25}{2}(2(-0.5)+(25-1)3.5)$

$S_{25}=1037.5$

thus the sum of 25 terms is 1037.5