Question

in an A. P 19th term is 52 and 38th term is 128, find sum of first 56 terms.

Answer 1

a+18d=52
a+37d=128
______________
19d=76, d=76/19 = 4
a= -20

then sum of 56terms
=(2(-20)+55(4))56/2
=180×28
=5040

Answer 2

Hii friend,

19th term of an AP = 52

A + 18 D = 52......(1)

And,

38th term of an AP = 128

A + 37D = 128.....(2)

From equation 1 we get,

A + 18D = 52

A = 52-18D....(3


Putting the value of A in equation (2)

A +37D = 128

52 - 18D +37D = 128

19D = 128-52

19D = 76

D = 76/19

D = 4

Putting the value of D in equation (3)

A = 52-18D

A = 52 - 18 × 4 = 52 - 72 = -20

First term of an AP = -20

Second term = a +D = -20 + 4 = -16

Third term = a+2D = -20 + 2 × 4 = -20 + 8 = -12

Sn = N/2 × [ 2A + (N-1) × D]

S56 = 56/2 × [ 2× -20 + (56-1) × 4]

=> 56/2 × (-40 + 220)

=> 56/2 × 180

=> 56 × 90 = 5040.

Hence,

The sum of 56 term of AP = 5040.

HOPE IT WILL HELP YOU.... :-)