Question

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

Answer 1

Answer:

a19=52

=>a+18d=52....(1)

a38=128

=a+37d=128....(2)

Subtracting (1) from (2)

a+37d-a-18d=128-52

=>19d=76

=>d=4

From (1)

a=52-18×4= -20

S56= 56/2{2×(-20)+(56-1)4}

=28(-40+220)

=28×180

=5040

Answer 2

Answer:

$\huge\bold\red{</strong><strong>➢</strong><strong><u>ANSWER</u></strong><strong><u> </u></strong><strong>:</strong><strong>-</strong><strong>}$

➢For an A.P., let a be the first term and d be the common difference.

==>> t19 = 52, t38 = 128 …[Given]

==>>Since, tn = a + (n – 1)d

∴ t19 = a + (19 – 1)d

∴ 52 = a + 18d i.e. a + 18d = 52 …(i)

Also, t38 = a + (38 – 1)d

∴ 128 = a + 37d i.e. a + 37d = 128 …(ii)

➢ Adding equations (i) and (ii), we get

==>> a + 18d = 52

(a + 37d = 128)/(2a + 55d = 180) .....(iii)

➢ Now, Sn = n/2 [ 2a + (n - 1)d]

∴ S56 = 56/2 [ 2a + (56 - 1)d]

= 28(2a + 55d) = 28 x 180 ....[From (iii) ]

∴ S56 = 5040

∴ The sum of the first 56 terms is 5040.