Question

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106 find the sum of 25 terms.​

Answer 1

Answer:

SOLUTION

Given that,

a3 = 12

a50 = 106

We know that,

an = a + (n − 1) d

a3 = a + (3 − 1) d

12 = a + 2d (I)

Similarly, a50 = a + (50 − 1) d

106 = a + 49d (II)

On subtracting (I) from (II), we obtain

94 = 47d

d = 2

From equation (I), we obtain

12 = a + 2 (2)

a = 12 − 4 = 8

a29 = a + (29 − 1) d

a29 = 8 + (28)2

a29 = 8 + 56 = 64

Therefore, 29th term is 64.

Answer 2

Answer:

$58$

Explanation:

a + 2d = 12

a + 49d = 106

47d = 94

d = 2

a + 4 = 12

a = 8

25 Th term:-

a + 24 D = 8 + 50 = 58