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

1.	In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.
Answer» For an A.P., let a be the first term and d be the common difference. t₁₉ = 52, t₃₈ = 128 …[Given] Since, t_n = 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, S_n = 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.

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

Answer»

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

t₁₉ = 52, t₃₈ = 128 …[Given]

Since, t_n = 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, S_n = 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.