Question

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Answer 1

Step-by-step explanation:

Solution - 4

We are given that the 2nd term of an A.P. = 7

→ a + (n - 1)d = 7

→ a + (2 - 1)d = 7

→ a + d = 7 ___ (1)

We are also given that the 4th term of an A.P. =23

→ a + (n - 1)d = 23

→ a + (4 - 1)d = 23

→ a + 3d = 23 ___ (2)

Subtracting equation (1) from (2)

→ a + 3d - (a + d) = 23 - 7

→ a + 3d - a - d = 16

→ 2d = 16

→ d = 16/2

→ d = 8

Therefore , common difference d = 8

Substituting the value of d in equation (1)

→ a + 8 = 7

→ a = 7 - 8

→ a = -1

Therefore, first term of an A.P. a = - 1

Now , finding 3rd and 5th term of the A.P.

a + (n - 1)d

= -1 + (3 - 1)8 = -1 + 2(8) = -1 + 16 = 15

a + (n - 1)d

= -1 + (5 - 1)8 = -1 + 4(8) = -1 + 32 = 31

Therefore, value of

→ a = -1

→ b = 15

→ c = 31