Question

In an Ap 19th term is 52 and 38th term is 128, find sum of first 56 terms

Answer 1

$Heya$‼

Given that,

19th term = 52
=> a + 18d = 52 ...(1)

and

38th term = 128
=> a + 37d = 128 ...(2)

Subtracting (1) from (2) ,
a + 37d = 128
a + 18d = 52
___________
0 + 19d = 76

=> $d=\frac{76}{19}$

=> $d=4$


$Substituting\:d = 4\:in\:(1) ,$
$a + 18(4) = 52$
$a + 72 = 52$
$a = 52 - 72$
$a = -20$


$Sum\:of\:56\:term:-$
$S^n=\frac{n}{2} [2a+(n-1)d]$
$S^{56}=\frac{56}{2} [2(-20)+(56-1)4]$
$S^{56}=28(-40+55*4)$
$S^{56}=28(-40+220)$
$S^{56}=28*180$
$S^{56}=5040$


Hence, the sum of 56 terms is 5040.

Answer 2

a19 = a+ 18d = 52........... 1
a38 =a+ 37d = 128..........2
- - -
= - 19d =-76
d = - 76/-19
d = 4
Putting this value of d in 1
a + 18 × 4 = 52
a + 72 = 52
a = 52 - 72
a= - 20
So
S56 = n/2 (2a +(n-1)d)
= 56/2 (2×-20 +(56 - 1) 4)
=28 (-40 + 55×4)
=28 (180)
=5040