Question

Find the sum of the first 56 terms of the AP
whose 4th term is 25 and 6 th term is 37.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Given→ a4 = 25

a6 = 37

.

a + 3d = 25 --------(1)

a + 5d = 37 --------(2)

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{ Using elimination method }

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a + 3d = 25

+a + 5d = 37

- - = -

___________

-2d = -12

d = 6

.

now put d=6 in equation (1)

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a + 3(6) = 25

a + 18 = 25

a = 7

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sum of 56 terms =

s = n/2 [ 2a + (n-1) d]

s = 56/2 [ 2(7) + (56 - 1)7 ]

s = 28 [ 14 + 55(7) ]

s = 28 [ 14 + 385]

s = 28 [399]

s = 11,172

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