Question

the first and last term of an AP are 2 and 125 respectively if the fifth term is 14 and the number of term in the Ap .​

Answer 1

Answer:

40

Step-by-step explanation:

14=2+(5-1)xd. or, d=3

Again,

125=2+(n-1)x3 or, n-1=41 or, n=40

where, d is common difference

n is nos. of elements/numbers in series.

Answer 2

First term, a1 = 2

Last term, an = 125

Fifth term, a5 = 14

d = ?

n = ?

a5 = 14

a + (n - 1)d = 14

2 + (5 - 1) d = 14

2 + 4d = 14

4d = 12

d = 12/4

d = 3

Now,

an = a + (n - 1)d

125 = 2 + (n - 1) 3

125- 2 = (n - 1)3

123/3 = n - 1

41 = n-1

41 + 1 = n

42 = n

Therefore, the number of terms, n is 42. (or The 42nd term is 125)