Question

find the sum of the first 25 term of an ap whose second term is 49 and fourth term is 21​

Answer 1

Hope its helpful...

Step-by-step explanation:

let first term of ap = a

common difference = d

now, 2nd term = 49 (given)

   4th term = 21

since  nth term =   a + (n - 1)d

so 2nd term = a +(2-1)d

49 = a + d       ...... 1

4th term = a + (4-1)d

21 = a + 3d       ........2

subtracting equation 1 from 2

a + 3d = 21

a + d   = 49

-   -           -

2d  = - 28

2d = -28

d = -28/2

d= -14

substituting value of d in ......1

a + d = 49

a + (-14) = 49

a = 49 + 14

a = 63

now to find sum of 1st 25 terms

sum of nth term = n/2 [ 2a + ( n-1 ) d ]

therefore, sum of 1st  25 terms =

25/2 [2.63 + (25 -1) -14]

= 25/2 [ 126 + 24 x (-14) ]

= 25/2 [ 126 - 336]

= 25/2 x ( -210 )

= 25 x (-105)

= -2625