Question

if the 5th of an ap is 31 and 25th term is 140 more than the 5th term find the AP

Answer 1

5th term = 31
a+(5-1)d=31
a+4d=31. .........1

25th term = 140 + 5th term
a+(25-1)d=140+31
a+24d=171. .........2

1-2
a+4d - (a+24d) =31-171
a+4d-a-24d=-140
-20d=-140
d=7

Put d=7 in 1
a+4d=31
a+4×7=31
a+28=31
a=31-28
a=3

The given A. P. is
a,a+d,a+2d,a+3d,.......
=3,3+7,3+2×7,3+3×7,.........
=3,10,17,24,......

Answer 2

Step-by-step explanation:

According to the Question

a = 31

⁵

i.e. a+4d = 31---------(i)

a = a + 140 [Given]

²⁵ ⁵

and,

a = a +24d

²⁵

Now, Put the value from equation(i)

a = 31 + 140

²⁵

Therefore, a = 171

²⁵

Then,

171 = a +24d ---------(ii)

$ubtract eq(i) from (ii), we get

20d = 140

Thus, d = 7

now, put this value in eq. (i)

So,

a + 4(7) = 31

a = 31 - 28

Thus, a = 3

Hence, The required A.P. is

3, 10, 17, 24, 31....