Question

Obtain the sum of the first 56 terms of an A.P. whose 19th and 38th terms are 52 and 128 respectively

Answer 1

Answer:

The required sum of 56 terms is 4816.

Step-by-step explanation:

nth term of an A.P is given by,

$\bf a_n = a+(n-1) d$

where

a denotes the first term

d denotes the common difference

Given,

19th term = 52

a + (19 – 1)d = 52

a + 18d = 52

a = 52 – 18d

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38th term = 128

a + (38 – 1)d = 128

a + 37d = 128

52 – 18d + 37d = 128

52 + 19d = 128

19d = 128 – 52

19d = 76

d = 76/19

d = 4

Therefore, common difference = 4

Finding the first term,

a = 52 – 18d

a = 52 – 18(4)

a = 52 – 72

a = –20

First term = –20

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Sum of first n terms of an A.P is given by,

$\bf S_n = \dfrac{n}{2} [2a+(n-1)d]$

Put n = 56,

$\sf S_{56} = \dfrac{56}{2}[2(-20) +(56-1) (4) ] \\\\ \sf S_{56}=28[-40+53(4)] \\\\ \sf S_{56}= 28[-40+212] \\\\ \sf S_{56}= 28(172) \\\\ \sf S_{56}=4816$

The required sum of 56 terms is 4816.