In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 term
Answers
Answer:
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. ⚡
Answer:
A19=a+18d=52
A38=a+37d=128
37d-18d=128-52
19d=76
d=4
a+18d=52
a=52-18×4
a=52-72
a=-30
S56= 28(-60+220)
= 28×160
=4480
Step-by-step explanation: