Question

An A.P. consist of50 terms.3

rd term is 12 and last term is 106 find 28

the term.find

sum of first 12 terms.​

Answer 1

Step-by-step explanation:

total number of terms 50

3rd term = a3 = a + 2d = 12

50th term = a50 = a + 49d = 106

subtract 3rd term from 50th term

a + 49d = 106

-(a + 2d = 12)

we get

47d = 94

d = 94/47

d = 2

substitute d = 2 in 3rd term a + 2d = 12

a + 2(2) = 12

a + 4 = 12

a = 12 - 4

a = 8

the first term of the AP is 8 and common difference is 2

28th term = a + 27d

= 8 + 27(2)

= 8 + 54

= 62

the 28th term of the AP is 62

sum of first 12 terms

Sn = n/2 (2a + (n-1)d)

$s12 = \frac{12}{2} (2(8) + (12 - 1)(2)) \\ = \frac{12}{2} (16 + (11 \times 2)) \\ = 6(16 + 22) \\ = 6 \times 38 \\ = 228$

the sum of first 12 terms of the AP is = 228

Hope you get your answer